Gamma Ray burst and things that blow up in the night

[derekmcd]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Well, that wasn't really my intent.&nbsp; It's just that neutron stars are thought to have a strong magnetic field, and galaxy centers are known to have "jets" too.&nbsp;&nbsp;&nbsp;&nbsp; Whatever the merger process of a galactic center might look like, it could in fact include "jets" since both the central masses can also produce "jets".&nbsp; Irrespective of what you think the center might contain, it seem seems as though collimated jets are possible. <br /> Posted by michaelmozina</DIV></p><p>Jets produced by AGN are via a steady inflow of material through the accrection disk.&nbsp; Nothing similar about a gamma ray burst through the merger of two neutron stars.&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>

 

[job1207]

BODY {font-family="Times New Roman"} TT {font-family="Courier New"} BLOCKQUOTE.CITE {padding-left:0.5em; margin-left:0; margin-right:0; margin-top:0; margin-bottom:0; border-left:"solid 2";} <table border="0" cellspacing="0" cellpadding="0" width="100%" bgcolor="#ece9d8"> <tbody> <tr bgcolor="#ece9d8"> <td> <div>From: Wayne Landsman <Wayne.B.Landsman@nasa.gov></div> <div>To:</div>Subject: Re: [Swift #2075] We are still here

 

[derekmcd]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>BODY {font-family="Times New Roman"} TT {font-family="Courier New"} BLOCKQUOTE.CITE {padding-left:0.5em; margin-left:0; margin-right:0; margin-top:0; margin-bottom:0; border-left:"solid 2";} From: Wayne Landsman <Wayne.B.Landsman@nasa.gov> To:Subject: Re: [Swift #2075] We are still here Hello, &nbsp;&nbsp;&nbsp; Your comment that "light decreases in intensity with the inverse square of the distance" is the correct one.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;__________________________________Enough said about that...!&nbsp; <br /> Posted by job1207</DIV></p><p>I'm not sure what your point is in repeating a previous post.&nbsp; I'm not saying the answer you recieved in an incorrect answer as I don't know the context of the question asked.</p><p>However, I am fairly confident in stating that a collimated beam of light does not follow the inverse square law.&nbsp; The energy is focused in a particular direction.&nbsp; The energy from the source of a collimated beam will not drop off in accordance with the ISL as the path of the photons are more parallel.&nbsp; The more collimated the beam, the less the intensity drops off over distance.&nbsp; The inverse square law applies to a point source that radiates in all directions.</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>

 

[michaelmozina]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>&nbsp; Your comment that "light decreases in intensity with the inverse square of the distance" is the correct one.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;__________________________________Enough said about that...!&nbsp; <br /> Posted by job1207</DIV></p><p>Do I get a cookie? </p> <div class="Discussion_UserSignature"> It seems to be a natural consequence of our points of view to assume that the whole of space is filled with electrons and flying electric ions of all kinds. - Kristian Birkeland </div>

 

[michaelmozina]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>I'm not sure what your point is in repeating a previous post.&nbsp; I'm not saying the answer you recieved in an incorrect answer as I don't know the context of the question asked.However, I am fairly confident in stating that a collimated beam of light does not follow the inverse square law.&nbsp; The energy is focused in a particular direction.&nbsp; The energy from the source of a collimated beam will not drop off in accordance with the ISL as the path of the photons are more parallel.&nbsp; The more collimated the beam, the less the intensity drops off over distance.&nbsp; The inverse square law applies to a point source that radiates in all directions. <br /> Posted by derekmcd</DIV></p><p>Unless the collimated beam was 100 percent collimated, as in a laser, it will still disperse over distance.&nbsp; It will also interact with plasma in the IGM.&nbsp; Even if it's not following the inverse square law, it will still disperse over time.</p><p>I'm starting to wonder now if these two thread (CME and this one) aren't related in some way.&nbsp; Imagine the earth being hit with collimated cathode ray from such a "jet"?&nbsp; Would it not create a CME like event around the sun? </p> <div class="Discussion_UserSignature"> It seems to be a natural consequence of our points of view to assume that the whole of space is filled with electrons and flying electric ions of all kinds. - Kristian Birkeland </div>

 

[derekmcd]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Unless the collimated beam was 100 percent collimated, as in a laser, it will still disperse over distance.&nbsp; It will also interact with plasma in the IGM.&nbsp; Even if it's not following the inverse square law, it will still disperse over time.I'm starting to wonder now if these two thread (CME and this one) aren't related in some way.&nbsp; Imagine the earth being hit with collimated cathode ray from such a "jet"?&nbsp; Would it not create a CME like event around the sun? <br /> Posted by michaelmozina</DIV></p><p>I'm not saying the intensity wouldn't fall off over time.&nbsp; The drop off in intensity would only be determined by the opening angle of the beam... It would not follow the inverse square law.&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>

 

[michaelmozina]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>I'm not saying the intensity wouldn't fall off over time.&nbsp; The drop off in intensity would only be determined by the opening angle of the beam... It would not follow the inverse square law.&nbsp; <br /> Posted by derekmcd</DIV></p><p>I would concur that it would depend on the opening angle , and the amount collimation of the beam.&nbsp; Over that distance however, even a direct hit from such a "beam' seems to be less than fatal. &nbsp;&nbsp; I would imagine that over a enough distance, it starts to act more like a standard point source (follows the inverse square law) than a collimated beam.&nbsp; I guess it depnds on the thickness and opacity of the materials it traverses in space.&nbsp; I can see how the intensity might be much higher than a non directed beam. </p> <div class="Discussion_UserSignature"> It seems to be a natural consequence of our points of view to assume that the whole of space is filled with electrons and flying electric ions of all kinds. - Kristian Birkeland </div>

 

[job1207]

<p>Frankly, I would write to the NASA guy noted above if you want to discuss this, and figure out just how this works. He does it for a living. He summed up my question very succintly in his note to me.&nbsp; </p><p>&nbsp;</p>

 

[DrRocket]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>I could be wrong, but I don't believe the inverse square law applies to collimated beams of electromagnetic radiation.&nbsp; <br />Posted by derekmcd</DIV></p><p>It does apply in the far field,&nbsp;once you can conside the source as a point with some dispersion angle.&nbsp;&nbsp; It does not&nbsp; take very long for most sources to be reasonably modeled as a point.&nbsp; And a few billion light years will certainly do.&nbsp;&nbsp;The inverse square&nbsp;fails to apply only if the beam is perfectly collimnated, with a zero angle of dispersion -- it is a basically a result of what a solid angle is, or the simple fact that the surface area of a sphere or piece of a sphere increases as the square of the radius.</p><p>But nothing has a zero angle of dispersion, not even a laser.</p><p>BTW the way derekmcd -- please be careful here as you are making me agree with Michael, a most uncomfortable position. ;-)</p> <div class="Discussion_UserSignature"> </div>

 

[derekmcd]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>It does apply in the far field,&nbsp;once you can conside the source as a point with some dispersion angle.&nbsp;&nbsp; It does not&nbsp; take very long for most sources to be reasonably modeled as a point.&nbsp; And a few billion light years will certainly do.&nbsp;&nbsp;The inverse square&nbsp;fails to apply only if the beam is perfectly collimnated, with a zero angle of dispersion -- it is a basically a result of what a solid angle is, or the simple fact that the surface area of a sphere or piece of a sphere increases as the square of the radius.But nothing has a zero angle of dispersion, not even a laser.BTW the way derekmcd -- please be careful here as you are making me agree with Michael, a most uncomfortable position. ;-) <br /> Posted by DrRocket</DIV></p><p>I may be I'm overthinking this.&nbsp; I think the point I'm trying to make is that you can't extrapolate any useful information about the source using the ISL if it is collimated (I do understand that no beam is perfectly so).&nbsp; I also understand you can model distant object and their luminosity as point sources whether collimated or not.&nbsp; However, the issue I'm trying to point out is that if the total release of energy is known, you can't simply apply the inverse square law to know how much energy will be recieved unless you know the energy is dispersed equally in all directions.&nbsp;&nbsp;</p><p>If a hypernova produces 10^50 joules (just throwing out a number).&nbsp; You can only apply the inverse square law to determine how much energy is recieved if the energy is released in all directions.&nbsp; If the energy is directionally released, you first need to know the opening angle to determine the energy recieved.&nbsp; Once that angle is determined, only then can you apply the inverse square law.&nbsp;</p><p>I guess it sounds like I'm contradicting myself...</p><p>I think I just didn't explain my initial train of though well enough and am struggling with my verbage.&nbsp; </p><p>Personal life has been in a bit of turmoil lately and haven't had much time to think before I post. <img src="http://sitelife.space.com/ver1.0/content/scripts/tinymce/plugins/emotions/images/smiley-yell.gif" border="0" alt="Yell" title="Yell" /> </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>

 

[DrRocket]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>I may be I'm overthinking this.&nbsp; I think the point I'm trying to make is that you can't extrapolate any useful information about the source using the ISL if it is collimated (I do understand that no beam is perfectly so).&nbsp; I also understand you can model distant object and their luminosity as point sources whether collimated or not.&nbsp; However, the issue I'm trying to point out is that if the total release of energy is known, you can't simply apply the inverse square law to know how much energy will be recieved unless you know the energy is dispersed equally in all directions.&nbsp;&nbsp;If a hypernova produces 10^50 joules (just throwing out a number).&nbsp; You can only apply the inverse square law to determine how much energy is recieved if the energy is released in all directions.&nbsp; If the energy is directionally released, you first need to know the opening angle to determine the energy recieved.&nbsp; Once that angle is determined, only then can you apply the inverse square law.&nbsp;I guess it sounds like I'm contradicting myself...I think I just didn't explain my initial train of though well enough and am struggling with my verbage.&nbsp; Personal life has been in a bit of turmoil lately and haven't had much time to think before I post. <br />Posted by derekmcd</DIV></p><p>You don't need to know that the energy is dispersed equally in all directions.&nbsp; You only need to know that the area covered by a solid angle grows as the square of the distance.&nbsp; The only exception is the zero angle, a singular case.</p><p>All that is involved is that the "beam" spreads out to cover some fixed proportion of a spherical shell.&nbsp; It&nbsp; need not even be uniformly distributed over that solid angle, since you can look a the problem locally.&nbsp; It is just like the magnification of an image that occurs with a projector -- the farther away the screen the larger the image, but the dimmer the image also.</p><p>What might be confusing is the model of lasers and flashlights.&nbsp; If you had a perfect point source located precisely at the focus of a perfectly parabolic reflector, then the beam would be be perfectly collimated along the axis of the parabola of the reflector and there would be no dispersion.&nbsp; But this is the zero angle of dispersion case, and is not physically realizable since any deviation from perfection in any of the assumptions destroys the perfect collimation.</p> <div class="Discussion_UserSignature"> </div>

 

[michaelmozina]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>It does apply in the far field,&nbsp;once you can conside the source as a point with some dispersion angle.&nbsp;&nbsp; It does not&nbsp; take very long for most sources to be reasonably modeled as a point.&nbsp; And a few billion light years will certainly do.&nbsp;&nbsp;The inverse square&nbsp;fails to apply only if the beam is perfectly collimnated, with a zero angle of dispersion -- it is a basically a result of what a solid angle is, or the simple fact that the surface area of a sphere or piece of a sphere increases as the square of the radius.But nothing has a zero angle of dispersion, not even a laser.BTW the way derekmcd -- please be careful here as you are making me agree with Michael, a most uncomfortable position. ;-) <br /> Posted by DrRocket</DIV></p><p>LOL.&nbsp; Get used to it. </p> <div class="Discussion_UserSignature"> It seems to be a natural consequence of our points of view to assume that the whole of space is filled with electrons and flying electric ions of all kinds. - Kristian Birkeland </div>