Astronomy/Physics trivia

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DrRocket

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>That's nearly right: It will come back to you...(that's a clue, btw) <br />Posted by eburacum45</DIV></p><p>Apparently some other people are trying to answer this same question.</p><p>http://www.physlink.com/education/askexperts/ae197.cfm<br /></p> <div class="Discussion_UserSignature"> </div>
 
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DrRocket

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>That's nearly right: It will come back to you...(that's a clue, btw) <br />Posted by eburacum45</DIV></p><p>OK I know where you are going, but I cheated (did not matter since I already submitted my guess).&nbsp; But once the answer comes out, there ought to be a lively, and educational, debate on what temperature means and in what reference frame you ought to be measuring it.<br /></p> <div class="Discussion_UserSignature"> </div>
 
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DrRocket

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Refrigeration is refrigeration, whether it occurs naturally or artificially. <br />Posted by eburacum45</DIV><br />&nbsp;</p><p>But if you take as the definition of temperature the average kinetic energy of the molecules, then temperature is dependent on the reference frame.&nbsp; That is the difference between static temperature and stagnation temperature.</p> <div class="Discussion_UserSignature"> </div>
 
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derekmcd

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>&nbsp;An explanatin of the photoeledtric effect.&nbsp; He recdeived it in 1921 and that explanation was one of the things that initiated the development of quantum mechanics -- the implications of which Einstein never fully accepted.</p><p>Posted by DrRocket</DIV></p><p>I was really hoping someone would chime in with one of his relativity theories.</p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>
 
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derekmcd

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>That's nearly right: It will come back to you...(that's a clue, btw) <br /> Posted by eburacum45</DIV></p><p>I had to cheat to find the answer, and I must say that clue is quite fitting.&nbsp;</p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>
 
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MeteorWayne

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>I had to cheat to find the answer, and I must say that clue is quite fitting.&nbsp; <br />Posted by derekmcd</DIV><br /><br />Might I suggest the next question (by the skilled winner) be posted in a new Astro/Phys Trivia Part 2 thread?</p><p>With the pain that it is to get to the current discussion, threads that don't run to multiple pages are MUCH easier to view.</p><p>Feel free to ignore my suggestion if needed... :)</p> <div class="Discussion_UserSignature"> <p><font color="#000080"><em><font color="#000000">But the Krell forgot one thing John. Monsters. Monsters from the Id.</font></em> </font></p><p><font color="#000080">I really, really, really, really miss the "first unread post" function</font><font color="#000080"> </font></p> </div>
 
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derekmcd

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Might I suggest the next question (by the skilled winner) be posted in a new Astro/Phys Trivia Part 2 thread?With the pain that it is to get to the current discussion, threads that don't run to multiple pages are MUCH easier to view.Feel free to ignore my suggestion if needed... :) <br /> Posted by MeteorWayne</DIV></p><p>I just thought this thread might be a fun way to get a little activity around here.&nbsp; I agree that if a question comes up that might spawn a lengthy discussion, it would be better to start a new thread.&nbsp; However, I think the questions themselves should all be contained in one thread to avoid bumping all the other threads off the first page. </p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>
 
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MeteorWayne

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Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>I just thought this thread might be a fun way to get a little activity around here.&nbsp; I agree that if a question comes up that might spawn a lengthy discussion, it would be better to start a new thread.&nbsp; However, I think the questions themselves should all be contained in one thread to avoid bumping all the other threads off the first page. <br />Posted by derekmcd</DIV><br /><br />I see your point. Just an idea to throw on the bonfire :) <div class="Discussion_UserSignature"> <p><font color="#000080"><em><font color="#000000">But the Krell forgot one thing John. Monsters. Monsters from the Id.</font></em> </font></p><p><font color="#000080">I really, really, really, really miss the "first unread post" function</font><font color="#000080"> </font></p> </div>
 
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DrRocket

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Refrigeration is refrigeration, whether it occurs naturally or artificially. <br />Posted by eburacum45</DIV></p><p>While we are waiting for someone to come up with the answer without cheating, I thought I might kick off the debate as to whether this really ought to be considered the lowest temperature in the universe.&nbsp; I will admit that I am not sure in my own mind whether or not this ought to be considered the lowest temperature.</p><p>Despite the references to reference frames, the treatment here is classical.&nbsp; I am assuming that the speeds involved are not relativistic.&nbsp; </p><p>What we have is a situation in which there is a gas at some pressure, contained in a volume the body of which we may consider to be at rest.&nbsp; The observer is considered to be in this same "at rest" reference frame.&nbsp; We will consider the Earth to be in the "at rest" frame. The gas in that body is cold, but not quite as cold as the background temperatue of the universe.&nbsp; Call that initial temperature T0.</p><p>The gas is&nbsp;released from confinement.&nbsp; It expands adiabatiatically and gains velocity as a result of that expansion.&nbsp; The temperature is sufficiently low that we can neglect any radiation from the gas or absorption by it.&nbsp; As a result of the expansion of the gas, the random motion of the molecules are converted into a bulk flow velocity plus a reduced level of random motion and molecular vibration (the combination of the random motion and molecular vibration is commonly called internal energy).&nbsp; Now, energy is conserved in this process so the average of the kinetic&nbsp; and potential energy in the random motions plus the average energy associated with the flow velocity is equal to the internal energy of the gas before it was released.&nbsp; In other words the average total energy of the gas, the internal energy plus the kinetic energy of flow remains, constant in the reference frame of the observer.&nbsp; Now temperature is defined to be proportional to average energy of the molecules, and that energy has not changed for&nbsp; the observer.&nbsp; In a reference frame moving with the flow at flow speed the kinetic energy of the gas molecules is less since the average velocity contains no component due to the flow velocity.&nbsp; The gas is clearly at temperature less than T0 in that moving reference frame and in this case is less than the cosmic background temperature.&nbsp; That is the temperatue you would measure from radiated emissions.&nbsp; But it is not the temperature that you would measure with a thermometer that is at rest.&nbsp; If the flow is brought to rest isentropically, say by putting an ideal thermometer, at rest, into the flow,&nbsp;you will recover the stagnation temperature, which is T0.</p><p>So, what do we count ?&nbsp; Temperature in our earthly reference frame or temperature in a moving reference frame ?</p><p>&nbsp;</p> <div class="Discussion_UserSignature"> </div>
 
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eburacum45

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<p>Since many people have googled it, I'll give the answer- the Boomerang Nebula. See http://apod.nasa.gov/apod/ap030220.html</p><p>We did discuss this nebula on the old forum, and I thought peopl;e would remember without googling (that's why I said 'it'll come back to you', apart from the obvious joke).</p><p>As far as the temperature goes- doesn't the Joule Thompson effect apply? http://en.wikipedia.org/wiki/Joule-Thomson_effect The temperature of the gas in this nebula is below the inversion temperature of hydrogen and helium, so they should cool, if that effect really does apply.</p> <div class="Discussion_UserSignature"> <p>---------------------------------------------------------------</p><p>http://orionsarm.com  http://thestarlark.blogspot.com/</p> </div>
 
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DrRocket

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Since many people have googled it, I'll give the answer- the Boomerang Nebula. See http://apod.nasa.gov/apod/ap030220.htmlWe did discuss this nebula on the old forum, and I thought peopl;e would remember without googling (that's why I said 'it'll come back to you', apart from the obvious joke).As far as the temperature goes- doesn't the Joule Thompson effect apply? http://en.wikipedia.org/wiki/Joule-Thomson_effect The temperature of the gas in this nebula is below the inversion temperature of hydrogen and helium, so they should cool, if that effect really does apply. <br />Posted by eburacum45</DIV></p><p>&nbsp;Yes it applies.&nbsp; That is precisely what I was describing, a reversible adiabatic expansion, aka an isentropic expansion.&nbsp; When that is done you basically are converting random motion of the gas molecules into a flow velocity, and when that is done the static temperature drops. But if you isentropically reduce the flow velocity back to zero you recover the original temperature, the stagnation temperature.&nbsp; The point here is that the temperature, which is proportional to the average kinetic energy of the molecules, depends on the reference frame of the observer.</p><p><br /><br />&nbsp;</p> <div class="Discussion_UserSignature"> </div>
 
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DrRocket

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<p>It has gotten awfully quiet in this thread and no one has bitten on the question of what temperature really means so let's try this one.</p><p>This theorem is a generalization to higher dimensions of the fundamental theorem of calculus.&nbsp; It is often seen by physicists or engineers in a course on vector analysis, electromagnetics or fluid dynamics, where it is an important tool.&nbsp; It is a famous theorem, though often seen in a more restrictive formulation with different notation.&nbsp; Here it is presented in modern notation.&nbsp; Name that theorem:</p><p><br /><img src="http://sitelife.space.com/ver1.0/Content/images/store/10/8/2a6fbede-85e9-44c4-b8c9-02effc4deb2d.Medium.gif" alt="" /></p> <div class="Discussion_UserSignature"> </div>
 
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