Equivalence between a Black Hole and travelling at the speed of light

In summary: There is no such thing as an "object moving at nearly the speed of light." In an inertial reference frame, light always travels at the speed of light. Any frame in which an object is moving at nearly the speed of light would be an incorrect frame of reference, and the laws of physics would not apply in that frame.
  • #36
JesseM said:
Gravitational time dilation in GR isn't normally symmetrical the way velocity-based time dilation in SR is...the lower clock will see the higher clock running faster than his own, while the higher clock will see the lower clock running slower than his own, and if they both use Schwarzschild coordinates they'll both agree the lower clock goes through fewer ticks than the higher one in any given interval of coordinate time.

Thanks JesseM. I need to read up about Schwarzschild coordinates and digest your answer. Where can I read more on GR time dilation not being symmetrical like velocity based time dilation?
 
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  • #37
DaleSpam said:
:rolleyes:If the wave is leaving a massive object then it is going up to a higher gravitational potential. This means that it is losing energy. Therefore the frequency becomes lower (redshift). Because the frequency is lower and c is constant the wavelength must be longer than when it was emitted.

Thanks DaleSpam. This may be basic for you, but this has really helped me. Is it also true that the wavelength from some atomic oscillation would be shorter if the mass from which it is being emitted is larger.
 
  • #38
Pierre007080 said:
Thanks JesseM. I need to read up about Schwarzschild coordinates and digest your answer. Where can I read more on GR time dilation not being symmetrical like velocity based time dilation?

Okay JesseM, youv'e scared me again. There is so much mathematics and terms like pseudo-Riemannian stuff that there is no way that I can see the light about why there is not symmetry in GR time dilation. Surely the radius from the centre of the mass is the main criterion as to the extent of GR time dilation and rod shortening? Spherical symmetry also sound sensible ... the concept of nested spheres satisfying Einsteins field equations sounds vaguely sensible, but I'm sure a more simplistic explanation is possible for us normal people? Perhaps the SR analogy of time dilation and rod shortening IS appropriate as an APPROXIMATION of the GR effects near of massive objects??
 
  • #39
DaleSpam said:
:rolleyes:If the wave is leaving a massive object then it is going up to a higher gravitational potential. This means that it is losing energy. Therefore the frequency becomes lower (redshift). Because the frequency is lower and c is constant the wavelength must be longer than when it was emitted.

If I understand this correctly a continuously emitting EM waveset would exhibit a short wavelength near the mass and an increasingly longer wavelength further away? This would surely be spherically valid at a set radius?
 
  • #40
Pierre007080 said:
Is it also true that the wavelength from some atomic oscillation would be shorter if the mass from which it is being emitted is larger.
Are you thinking of something like a spring-mass system at an atomic level, or are you thinking more about something like the hyperfine transition that defines an atomic clock.

Pierre007080 said:
If I understand this correctly a continuously emitting EM waveset would exhibit a short wavelength near the mass and an increasingly longer wavelength further away?
Yes.
 
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  • #41
DaleSpam said:
Are you thinking of something like a spring-mass system at an atomic level, or are you thinking more about something like the hyperfine transition that defines an atomic clock.

DaleSpam, you are going where I fear to tread. I was trying to find a way to ask what would happen if the mass increased and the emission was from the same source as it would have been when the mass were smaller. Perhaps black body radiation from two massive bodies that have co-incidently reached the same temperature?
 
  • #42
Blackbody radiation is a function of temperature only. It does not depend on the mass of the object nor on the mass of its constituents.

If you are trying to ask whether or not gravitational time dilation could be interpreted as increased mass, then I think the answer is "not in general". In other words, there may be some specific cases where you could (e.g. mass/spring system) but many cases where you could not (e.g. pendulum or blackbody).
 
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  • #43
DaleSpam said:
Blackbody radiation is a function of temperature only. It does not depend on the mass of the object nor on the mass of its constituents. If you are trying to ask whether or not gravitational time dilation could be interpreted as increased mass, then I think the answer is "not in general". In other words, there may be some specific cases where you could (e.g. mass/spring system) but many cases where you could not (e.g. pendulum or blackbody).
Would it or not be dependant on the gravitational altitude of its location?
 
  • #44
Austin0 said:
Would it or not be dependant on the gravitational altitude of its location?
Yes. Or rather, the spectrum received would depend on the temperature of the blackbody and the gravitational time dilation between the blackbody and the receiver. The time dilation then is dependant on the difference in "gravitational altitude".
 
  • #45
DaleSpam said:
Yes. Or rather, the spectrum received would depend on the temperature of the blackbody and the gravitational time dilation between the blackbody and the receiver. The time dilation then is dependant on the difference in "gravitational altitude".
Thanks

Is there any way to measure electron resonance frequencies other than by the frequency of light they can absorb??
 
  • #46
DaleSpam said:
Blackbody radiation is a function of temperature only. It does not depend on the mass of the object nor on the mass of its constituents.

I think that I put my question poorly. The fact that I referred to blackbody at the same temp was to get the same emission frequency from two massive objects of differing masses. The question is this: If we were to measure the wavelength of the emitted ray at the same distance from the centre of the two bodies, would the wavelength of the radiation from the more massive body be shorter?
 
  • #47
JesseM said:
Gravitational time dilation in GR isn't normally symmetrical the way velocity-based time dilation in SR is...the lower clock will see the higher clock running faster than his own, while the higher clock will see the lower clock running slower than his own, and if they both use Schwarzschild coordinates they'll both agree the lower clock goes through fewer ticks than the higher one in any given interval of coordinate time.

I think I understand. Symmetry is being used in a way that I'm unsure of. Do you mean that in SR the sum of velocities is different of that of GR. In other words the time dilation and rod shortening in GR are the same for any observer whereas those of SR are relative to the observer's motion?
 
  • #48
Pierre007080 said:
If we were to measure the wavelength of the emitted ray at the same distance from the centre of the two bodies, would the wavelength of the radiation from the more massive body be shorter?
No, the wavelength from the more massive body would be longer. I refer you back to post 35.

DaleSpam said:
If the wave is leaving a massive object then it is going up to a higher gravitational potential. This means that it is losing energy. Therefore the frequency becomes lower (redshift). Because the frequency is lower and c is constant the wavelength must be longer than when it was emitted.
For the more massive body the change in gravitational potential is greater so the redshift is greater so the frequency is lower so the wavelength is longer.
 
  • #49
DaleSpam said:
No, the wavelength from the more massive body would be longer. I refer you back to post 35.

For the more massive body the change in gravitational potential is greater so the redshift is greater so the frequency is lower so the wavelength is longer.

This seems like a contradiction to me. Please bear with my persistence as this is important to my understanding of GR. May I restate the question in the following way: If we were on two massive bodies of exactly the same size (radius) but different masses and we're standing on the surface. Would the black body radiation emitted radially (away from the centre of the mass) from two similar items be emitted at the same wavelength and frequency?
 
  • #50
Pierre007080 said:
If we were on two massive bodies of exactly the same size (radius) but different masses and we're standing on the surface. Would the black body radiation emitted radially (away from the centre of the mass) from two similar items be emitted at the same wavelength and frequency?
This is going in circles. Why don't you try to answer this question (frequency of emitted radiation) with what I have already provided (see post 42). Then try to answer the question about what would the frequency of the received radiation be (see post 35) with the hopefully obvious stipulation that the receiver will be above the blackbody surface.
 
  • #51
DaleSpam said:
Yes. Or rather, the spectrum received would depend on the temperature of the blackbody and the gravitational time dilation between the blackbody and the receiver. The time dilation then is dependant on the difference in "gravitational altitude".

Thanks for your response. I have re-read the previous discussion but the discrepancy seems to be that I am under the impression that time (the tick of the clock) exists with regard to the "gravitational potential". The "tick of the clock" surely has to be slower at the higher "gravitational potential" on the surface of the larger mass? The "dilation" depends upon the movement of the wave to a higher "gravitational potential". The reason I chose the surface of two same size objects as my reference frame was to eliminate the movement (or dilation) aspect and focus on the fact that the larger mass has a HIGHER gravitational potential at it's surface and that the EMISSION frequency and wavelength would differ as observed by a distant observer at right angles to the emission, even though the temperature of the body is the same. Of course both local parties would measure the frequency as that typical of the black body radiation spectrum.
 
  • #52
Pierre007080 said:
the larger mass has a HIGHER gravitational potential at it's surface
No, the larger mass has a lower gravitational potential (more negative) at its surface. This is the case in Newtonian gravity as well as in the Schwarzschild metric in GR.

Pierre007080 said:
the EMISSION frequency and wavelength would differ as observed by a distant observer
This is self-contradictory. Only a local observer can measure the emission frequency. (Actually, even a local observer technically still measures a received frequency, but there are no relative Doppler or time dilation effects for a local observer so a local observer always receives the same frequency as emitted.)
 
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  • #53
DaleSpam said:
No, the larger mass has a lower gravitational potential (more negative) at its surface. This is the case in Newtonian gravity as well as in the Schwarzschild metric in GR.

My mistake. The gravitational potential is "lower" at the surface of the more massive body. Will a clock not tick slower near the large mass?


This is self-contradictory. Only a local observer can measure the emission frequency. (Actually, even a local observer technically still measures a received frequency, but there are no relative Doppler or time dilation effects for a local observer so a local observer always receives the same frequency as emitted.)

I understand the local observer point of view, but why should there not be a distant observer at a different gravitational potential viewing the wave emission from the side?
 
  • #54
Pierre007080 said:
why should there not be a distant observer at a different gravitational potential viewing the wave emission from the side?
Certainly you can have such an observer. That would not change anything I said above.
 
  • #55
JesseM said:
Gravitational time dilation in GR isn't normally symmetrical the way velocity-based time dilation in SR is...the lower clock will see the higher clock running faster than his own, while the higher clock will see the lower clock running slower than his own, and if they both use Schwarzschild coordinates they'll both agree the lower clock goes through fewer ticks than the higher one in any given interval of coordinate time.

From JesseM's explanation I understood that a distant observer would be able to tell that the lower (Lower gravitational potential) observer would have a clock ticking slower?
 
  • #56
Pierre007080 said:
From JesseM's explanation I understood that a distant observer would be able to tell that the lower (Lower gravitational potential) observer would have a clock ticking slower?
Yes.
 
  • #57
DaleSpam said:
Yes.

Why then could a third observer at right angles to the waveset not view both the first two observers (at the low and high gravitational potentials) and recognise that they are experiencing different wavelengths of the same waveset?
 
  • #58
I don't know what you mean by that. What is a waveset? What does it mean to experience a wavelength of a waveset?
 
  • #59
Originally Posted by Pierre007080
If I understand this correctly a continuously emitting EM waveset would exhibit a short wavelength near the mass and an increasingly longer wavelength further away?

Yes.[/QUOTE]
 
  • #60
This is self-contradictory. Only a local observer can measure the emission frequency. (Actually, even a local observer technically still measures a received frequency, but there are no relative Doppler or time dilation effects for a local observer so a local observer always receives the same frequency as emitted.)[/QUOTE]

Hi DaleSpam, It is this previous quote of yours that I am trying to get my head around by suggesting a third observer. I can't understand how you can state that only a local observer can measure the emmission frequency.

Would a third observer not measure a DIFFERENT emission frequency and spectrum from the same temperature black body on the surface of two bodies of different mass because of the different "gravitational potential".
 
  • #61
OK, let's simplify things. Instead of talking about measuring an emitted frequency let's only talk about emitting or receiving a frequency. Let's stipulate that you can perfectly determine the emitted frequency (e.g. using an ideal clock and an ideal waveform synthesizer) and that you can perfectly determine the received frequency (e.g. with an ideal noise free detector and an ideal clock).

Can you re-ask your question in those terms.
 
  • #62
DaleSpam said:
OK, let's simplify things. Instead of talking about measuring an emitted frequency let's only talk about emitting or receiving a frequency. Let's stipulate that you can perfectly determine the emitted frequency (e.g. using an ideal clock and an ideal waveform synthesizer) and that you can perfectly determine the received frequency (e.g. with an ideal noise free detector and an ideal clock).

Can you re-ask your question in those terms.

Hi DaleSpam. Thanks for your attempts to help me. My interest is in cosmology and I was seeking a simple UNCOMPLICATED answer about emission from these masses and the gravitational potential within which the occur. The conclusions regarding galactic rotation curves etc made from observations still don't make sense to me, but I have decided to back off from trying to extract a "rule of thumb" regarding how GR affects our observations. As you say, we seem to be going in circles. I thank you for your patience. Regards. Pierre.
 
  • #63
Pierre007080 said:
Hi DaleSpam. Thanks for your attempts to help me. My interest is in cosmology and I was seeking a simple UNCOMPLICATED answer about emission from these masses and the gravitational potential within which the occur. The conclusions regarding galactic rotation curves etc made from observations still don't make sense to me, but I have decided to back off from trying to extract a "rule of thumb" regarding how GR affects our observations. As you say, we seem to be going in circles. I thank you for your patience. Regards. Pierre.
You are quite welcome. For an uncomplicated answer I would just stick with post #35, everything else is just window-dressing and confusion.
 

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