Relativity question

[mcbethcg]

OK, here's a question.<br /><br />I have been told that when an object attains a high enough velocity, it gains relativistic mass, and that relativistic mass is different from regular mass in that it 1) does not create gravity and<br />2) resists acceleration and decelleration in the direction of movement, but not in changes of direction.<br /><br />This leads me to a thought experiment.<br />I don't believe in reactionless drives, so don't think I am a wing-nut. I just think what I was told about relativistic mass was wrong.<br /><br />Here is the thought experiment.<br />Have a particle accelerator accelerate particles to 99.99% of the speed of light. Send the beam around a bent track, that curves 90 degrees, and goes into a target that stops the beam.<br /><br />In non-relativistic physics, the push on the ends of the track created by accelerating and decellerating the particles is exactly countered by the centrepital force created by changing the direction of the beam. With relativistic mass, though, what's up?

 

[newtonian]

mcbethcg- Others will answer better. I'm not even sure.<br /><br />If I am correct, the rest mass is the mass by which you would do calculations such as gravity.<br /><br />The relativistic mass is merely a matter of appearance from a different reference point.<br /><br />In other words, the particle (or object, star, planet, etc.) is not moving in reference to itself, but is moving in reference to another point.<br /><br />I dare not go further with this - I will listen for more knowledgeable posters.

 

[rogers_buck]

Humm. Not too sure what your getting at, so I'll blather a bit and you can jump in and set me straight.<br /><br />Your talking about mass energy equivalence in General Relativity, specifically good ole E=mc^2.<br /><br />The mass change is equivalent to delta Ec^-2 and the energy of the reference frame (in your examples case is its kinetic energy.<br /><br />You said one thing I need to explore more. The relativistic mass is relative to the reference frame of the moving particle. In terms of the rest reference frame nothing has changed with respect to the mass of the particle. So bending magnets at the turn should be able to change the direction of the beam just the same as if the particles were traveling without relativistic effects. But the electric field of the charged particle at relativistic velocities would act as if it were in a gravitational well and would not be invariant with respect to the Lorentz force. So your magnetic field would require a strong enough field density to deflect the relativistic mass.<br /><br />When your particles hit the target the actual effective mass will not be the particle mass+target as the collision will radiate energy and thereby change the effective mass (again delta E*c*2).<br /><br />So in effect, the relativistic system is balanced just as the Newtonian system. Is this what you meant?<br />

 

[siarad]

A change in direction certainly is dependent upon relativistic mass. A correction has to be made for the increased difficulty of deflecting the electrons travelling down an oscilloscope tube above 450 million mph

 

[mcbethcg]

Thanks guys. I assumed that change in direction was dependent on relativistic mass, but had been told it was not. Just getting a second opinion.

 

[rogers_buck]

Not exactly. If your particle had a little rocket motor on it that was used to make the turn, the rocket motor would only have to compensate for the newtonian mass of the particle so long as it did not accelerate or decelerate the particle. This is equivalence in action. <br /><br />Your problem is that you are effecting the particle from the outside. That is why you have to overcome its well with the Lorentz adjustment, and that is why this is equivalent to the relativistic mass of the particle.<br /><br />Not quite the same thing, but your friend wasn't pulling your leg. Reference frame does matter.<br />

 

[control_group]

OK, first off, my understandings/assumptions, so I can be corrected if need be.<br /><br />1) As velocity approaches <i>c</i> mass increases (also length decreases and time dilates, but it's the mass increase that's causing me problems)<br /><br />2) Assuming a non-rotational inertial frame, there is no difference between the statements "I am moving towards the item" and "the item is moving towards me."<br /><br />If those two statements are correct, then I've got this question: say a person has accelerated up to some giant fraction of <i>c</i>, such that the mass increase effect is significant. He's hurtling through space, but from his frame, space is hurtling past him. At some point, something (say a hydrogen atom) is in his path.<br /><br />On the one hand, he shouldn't be able to tell whether he or the hydrogen atom is "moving." But, if he were to measure the mass of the hydrogen atom (gravitometrically, say), wouldn't he be able to tell that it wasn't more massive than it should be, and therefore conclude that his frame was <i>not</i> a rest frame?

 

[vandivx]

"On the one hand, he shouldn't be able to tell whether he or the hydrogen atom is "moving." But, if he were to measure the mass of the hydrogen atom (gravitometrically, say), wouldn't he be able to tell that it wasn't more massive than it should be, and therefore conclude that his frame was not a rest frame?"<br />------------------------<br />yeah, relativity is not all that 'relative' after all, welcome to the club of those able to think critically, that is not as sheep do exhibit this behavior<br /><br />however physicists know well why they play sheep, that's because dealing with absolute space is pesky problem and its more convenient to play it that way<br /><br />vanDivX <div class="Discussion_UserSignature"> </div>

 

[why06]

Im not sure why the mass increases....<br /><font color="yellow"> Perhaps it is that the same mass takes up more space-time therefore causing a bigger depresion in the fabric of space hence we would assume the mass has increased.<font color="white"><br /><br /><br />"..Does a car have greater mass traveling at 30 mph than at 0mph...?" <br />- one of the experts should be along in a sec to answer this <img src="/images/icons/wink.gif" /><br /><br />Good Luck, <img src="/images/icons/smile.gif" /></font></font> <div class="Discussion_UserSignature"> <div>________________________________________ <br /></div><div><ul><li><font color="#008000"><em>your move...</em></font></li></ul></div> </div>

 

[R1]

I think the length increases,<br />But either way, then couldn't the observer watch for a hydrogen string<br />instead of a hydrogen atom?<br /> <div class="Discussion_UserSignature"> </div>

 

[toothferry]

Smart eye, but wrong terms being mixed. <img src="/images/icons/smile.gif" /><br /><br />The "Mass" doesn't change according to Relativity, according to the "Mass" that you are referring to. ["invariant mass (also known as the rest mass, intrinsic mass or the proper mass) is an observer-independent quantity"] <br /><br />There needs to be another term, but Relativity refers to Inertial Mass or Relativistic Mass (google those terms).<br /><br />http://en.wikipedia.org/wiki/Relativistic_mass

 

[ihwip]

If a star is moving away from Earth and photons emitted from the star are reaching Earth shifted red and thus have reduced energy, where did the missing energy go? As far as I know photons have a set mass. This would mean energy is going somewhere.<br /><br />Is it more of a 'stream' of energy? Where the missing energy is creating extra photons coming from the star as the energy is stretched out? That really doesn't make sense but that is all I can think of.

 

[jgreimer]

If someone throws a baseball at you while you are running away from them, you don't catch the baseball with the same force as you would if you were standing still. Where did that energy go?

 

[Mee_n_Mac]

This question came up in another thread not that long ago. I'll see if I can find it.<br /><br />EDIT : Here's the link to that thread here in AtA. Note speedy's answer contained therein. I'm still not sure I buy it but I sure don't have anything better to counter with.<br /><br /> <div class="Discussion_UserSignature"> <p>-----------------------------------------------------</p><p><font color="#ff0000">Ask not what your Forum Software can do do on you,</font></p><p><font color="#ff0000">Ask it to, please for the love of all that's Holy, <strong>STOP</strong> !</font></p> </div>

 

[ihwip]

A photon seems to keep its mass and velocity the same while its frequency is reduced. A baseball doesn't move relatively in that its observed speed can change. (The answer is that the energy goes into letting the ball keep up with you btw haha)<br /><br />I am just thinking that this might be a clue as to how relativity works and I am really interested in the answer. Thanks for the quick responses.

 

[jgreimer]

No, the baseball's energy isn't reduced at all. It just appears that way from your perspective as a moving observer. The same ball caught by someone else with a different motion relative to the baseball would measure the baseball's energy differently.<br /><br />Likewise from the perspective of someone at the light source the light's wavelength hasn't changed nor has its energy. Its wavelength has changed only due to the different perspective of the moving observer.<br /><br />While this example is not a good one for demonstrating relativity, it does illustrate that the relative motions of two observers can influence their measurements.

 

[SpeedFreek]

Heh, that thread was an education Mee_n_Mac! But that was about redshift due to the expansion of space, rather than redshift due to relative inertial motion through space. Is the energy lost in the same way?<br /><br />I still think the answer is essentially the same either way though, as the wavelength increases so the frequency reduces and therefore nothing is actually lost, it just gets distributed along the length of the wave, which is now longer. But please don't as me to <i>quantify</i> that in terms of single photons!<br /><br />Remember the double split experiment and its variations.<br /><br />Treat light as a wave hitting two slits and it acts like a wave and produces an interference pattern on a plate on the other side.<br /><br />Treat light as a stream of individual photons hitting either of those slits and it still builds up those interference patterns on a detector over time.<br /><br />But as soon as you try to see which slit a photon went through, the interference pattern disappears. As soon as you start asking questions about specific photons the result changes! <img src="/images/icons/wink.gif" /> <div class="Discussion_UserSignature"> <p><font color="#ff0000">_______________________________________________<br /></font><font size="2"><em>SpeedFreek</em></font> </p> </div>

 

[Mee_n_Mac]

Wave-particle duality is among one of the most confounding "simple" things I've yet to wrap my brain around. As to the OP's question ... I guess I assumed it was the same as the other thread but rereading it, yes I can see it meaing proper motion rather than metric expansion of space. <div class="Discussion_UserSignature"> <p>-----------------------------------------------------</p><p><font color="#ff0000">Ask not what your Forum Software can do do on you,</font></p><p><font color="#ff0000">Ask it to, please for the love of all that's Holy, <strong>STOP</strong> !</font></p> </div>

 

[weeman]

jgreimer: <font color="yellow"> While this example is not a good one for demonstrating relativity, it does illustrate that the relative motions of two observers can influence their measurements. </font><br /><br />I think it's a great explanation within just a few sentences <img src="/images/icons/smile.gif" /><br /><br />IHWIP, as jgreimer said, it is all based on your relative position as an observer, hence the name, Relativity. The way I see it, in some ways, Einstein's ideas of relativity can be quite simple.<br /><br />As an observer on Earth, you are watching the star move farther away from you, wondering where the additional energy goes from the redshifted photons. However, if you were an observer on the star, you would observe Earth moving away from you and argue that Earth's light is redshifted, and state that Earth is moving away from <i>you</i>, not <i>you</i> moving away from Earth. <br /><br />So, this is where the paradox is settled based on the observer's frame of reference.<br /><br />I have a recommendation IHWIP, go out and buy Brian Greene's, 'The Elegant Universe'. The beginning of the book tells a great deal of Einstein's theories of Relativity. <div class="Discussion_UserSignature"> <p> </p><p><strong><font color="#ff0000">Techies: We do it in the dark. </font></strong></p><p><font color="#0000ff"><strong>"Put your hand on a stove for a minute and it seems like an hour. Sit with that special girl for an hour and it seems like a minute. That's relativity.</strong><strong>" -Albert Einstein </strong></font></p> </div>

 

[alokmohan]

Mtion depends on state of movement of observer.There is no absolute motion.

 

[origin]

I don't know where the energy goes.<br /><br />This is not right though:<br /><br /><font color="yellow">I still think the answer is essentially the same either way though, as the wavelength increases so the frequency reduces and therefore nothing is actually lost, it just gets distributed along the length of the wave, which is now longer. </font><br /><br />Frequency and wavelength are just two ways of measuring the same thing since the speed of light is constant. <br /><br />If blue light is shifted to red by the expansion of the universe then the photon does in fact have a lower energy.<br /><br />Energy of a photon = hc/wave length<br /><br />h = plancks constant<br />c = speed of light<br />Blue light 400 nm, Red light 700 nm<br /><br />If you chose to use frequency you will get the same energy drop:<br /><br />Energy of a photon = hf<br /><br />Blue light = 7 X 10^14 hertz <br />Red light = 4.3 X 10^14 hertz<br /><br />I think it is a good question - which I will look into if I ever get some time.... <br /><br />Edited to add, it is actually pretty obvious 'where the energy goes', I think someone already made the point and I just didn't 'get it'. I will discuss it more later. <div class="Discussion_UserSignature"> </div>

 

[origin]

jgreimer is right, and the baseball analogy is a good one. <br /><br />Lets assume that a baseball pitcher is standing on a flat car facing backwards on a train that is moving at 90 mph. If he somehow could time it, as he passes a stationary person on the side of the tracks and throws a 95 mile an hour fast ball to him in the opposite direction of travel, it will only be going 5 mph when the stationary guy catches it. <br />If there is a person on the other end of the flat car and the pitcher throws him a fast ball, it really will be going 95 mph. The same amount of energy was put into both pitches, but one individual sees a 95 mph pitch and the other sees a 5 mph pitch. <br /><br />Why? Because the resultant energy is relative to the COMBINED velocities both + and -. In other words for the stationary guy there is a large 'negative' kinetic energy in the ball when it is held by the pitcher moving away from him and when Pitcher pitches the ball he inputs a certain amount of 'positive' kinetic energy and when these are added together whatever is left over is the final KE for the stationary guy. <br /><br />Another way to look at this is the pain would be the same if you were beaned by a 90 mph fastball or if you had your head out of a car window going 90 mph and hit a stationary baseball. Even though the stationary baseball had no KE it will still hurt like hell. <br /><br />So back to the light problem. <br />Assume that an individual has a laser that emits blue light and he hops on a starship, points the laser back at you and accelerate to a high velocity away from you. You would see the light from the laser change color from blue through the spectrum to red because it is now at a lower energy. The laser photons are still moving at the speed of light because of course c is constant (which is the difference between this and the baseball analogy), but there is a decrease in energy (frequency), just like there was in the baseball example. <br /><br />So you might now say, "so wher <div class="Discussion_UserSignature"> </div>

 

[alokmohan]

Mtion depends on state of inertia of a thing.Am I correct to say?

 

[siarad]

<blockquote><font class="small">In reply to:</font><hr /><p>speed of light is constant<br /> since the velocity can't decrease<p><hr /></p></p></blockquote><br />Surely the <i>measured</i> speed of light is constant but it is <i>fixed</i> by the medium, it being impossible to measure incoming light's speed or outgoing for that matter, M&M experiment proved that.<br />Therefore the baseball analogy seems correct.<br />However the ratio of incoming light's speed to movement is tiny w.r.t. the baseball action.<br />Doesn't the redshift tell us the apparent reduction in the incident lights speed as with sound. Although, unlike light, we do have a way of ascertaining our speed w.r.t. sound, the Mach meter.

 

[nexium]

Lets say there is one watt of light energy from a supernova is illuminating all of Earth. That would be two watts if the object was not red shifted = receding. Because it is receeding; the super nova will be visable twice as long, thus compensating for the half as much. Neil