Faster than the speed of light

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  • #1
Jack
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If time slows down as you approach the speed of light then does this mean that it stops when you reach the speed of light and/or runs backwards above the speed of light?
 

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  • #2
chroot
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If those velocities were possible, yes, that's what would happen. If you were to go faster than light, you would be performal 'acausal' motion, and could go backwards in time and kill your father and so on.

It is true that there is a viable theory of particles called 'tachyons,' which always travel faster than light. The speed of light is a barrier -- anything traveling < c will always travel < c, and anything traveling > c will always travel > c. It's a brick wall in both directions.

Some number of experiments were designed, built, and run to detect such tachyons -- but none were ever found.

- Warren
 
  • #3
jeff
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Jack,

The speeds of objects with non-zero rest mass can approach but never equal light speed since then their relativistic mass would become infinite, violating conservation of energy.
 
  • #4
Jack
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Originally posted by chroot
It is true that there is a viable theory of particles called 'tachyons,' which always travel faster than light. The speed of light is a barrier -- anything traveling < c will always travel < c, and anything traveling > c will always travel > c. It's a brick wall in both directions.
- Warren

Tachyons were going to be the subject of my next question actually. Everything you said about them was correct although I don't know much more about them than that, in fact I don't think anyone knows much more about them. Does this mean that they travel backwards in time and what are the implications of this?

And stenitz, I was going to add that I realized faster than the speed of light travel was possible because I just knew that someone would say that.
 
  • #5
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Originally posted by Jack
Tachyons were going to be the subject of my next question actually. Everything you said about them was correct although I don't know much more about them than that, in fact I don't think anyone knows much more about them. Does this mean that they travel backwards in time and what are the implications of this?

The implications of traveling backward in time (according to GR), is that one is conducting negative travel through space. This makes no sense, of course, as there is no such thing as negative movement through all three of the spatial dimensions, at the same time (at least not that I can concieve of).
 
  • #6
Viper
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Its impssible to go faster than the spped of light
 
  • #7
marcus
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Originally posted by Viper
Its impssible to go faster than the spped of light

Viper, you and I are receding at twice the speed of light from the people (if there are any) living in a certain galaxy which was observed only last year.

Superluminal speeds of recession are part of General Relativity and routine in cosmology.

When you say no speeds can be faster than light be sure to specify that it is in the context of Special Relativity. It is true in that context and not true in the universe at large of General Relativity.

It causes terrible confusion when people say "no speeds faster than light" without being clear about the context. Physicists do the public a great dis-service when they fail to point that out.

People get the absurd notion that it is an absolute truth about nature and it interferes with their understanding of basic facts like the Hubble law and expansion.

This is also a remonstrance with some of the other posts in this thread which were insufficiently qualified as well.
 
  • #8
AndersHermansson
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Originally posted by marcus
Viper, you and I are receding at twice the speed of light from the people (if there are any) living in a certain galaxy which was observed only last year.

ALERT! CONTRADICTION DETECTED! :)
How could we have observed something that is receeding twice the speed of light?
 
  • #9
Hurkyl
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How could we have observed something that is receeding twice the speed of light?

Might I ask what by reason you think we shouldn't be able to see such a thing?
 
  • #10
marcus
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Originally posted by AndersHermansson
ALERT! CONTRADICTION DETECTED! :)
How could we have observed something that is receeding twice the speed of light?

Yes, Anders, please tell us why you think this is contradictory!
Hurkyl wants to know why you think this too.
The object was a quasar observed in 2002 at z = 6.4
The current estimated distance to the thing is 28 billion LY and the
current speed of recession is 2c.

http://www.sdss.org/news/releases/20030109.quasar.html

One does not observe the quasar as it is today, of course, since its light took over 12 billion years to get here. Because of the expansion that occurred during that time, its present distance is calculated to be, as I said, 28 billion LY. And its distance is increasing at the rate 2c.

Even though you think this is at odds with Special Relativity, it is not-----special relativity, with its speed limit, simply does not apply in this case so there is no logical contradiction.

But if you think there is, please explain it to us :wink:
 
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  • #11
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marcus, perhaps it would be appropriate to explain to Anders why this is possible.

Anders, there are a few reasons that something could be observed as moving faster than c. One is redshift, which you are probably already familiar with. Another is the fact that the expansion of space is taking place at a rate greater than c.
 
  • #12
Mentat
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Also, marcus and Hurkyl, Anders' questions appears to be slightly different than might be assumed. You see he was asking how we could observe something, if it were moving at a speed greater than c. He wasn't (IMO) asking how we could observe something as moving faster than c (even though that's what I initially assumed, also).

Perhaps AndersHermansson should explain it himself though, as the above is just the result of a "twice-over" by me.
 
  • #13
marcus
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Originally posted by Mentat
marcus, perhaps it would be appropriate to explain to Anders why this is possible.

Anders, there are a few reasons that something could be observed as moving faster than c. One is redshift, which you are probably already familiar with. Another is the fact that the expansion of space is taking place at a rate greater than c.

Mentat, I just got back to the computer and saw your post.
This is a helpful comment and clarifies things. Also what you said after that is relevant. I will respond momentarily.
 
  • #14
marcus
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Originally posted by Mentat
Also, marcus and Hurkyl, Anders' questions appears to be slightly different than might be assumed. You see he was asking how we could observe something, if it were moving at a speed greater than c. He wasn't (IMO) asking how we could observe something as moving faster than c (even though that's what I initially assumed, also).

Perhaps there are two separate questions, one that you just posed and one that Anders asked.

If you ask "how can we observe something in the act of moving at > c?"

then the answer is that we can't. Once something is receding at over the speed of light it can never send a signal that reaches us. Light from that object will never get here.

[self editing: sorry erase this. we can sometimes observe things receding at > c. Please disregard what I said which was not sufficiently thought out. See Hurkyl's immediately following comment.]

But I believe Anders was responding to an example I gave of an object which we can see today (as it was 13 billion years ago) and which is now calculated to be 28 billion LY away and receding at 2c-----this is this redshift 6.4 quasar observed last year.

I said:
--------------------------------------------------------------------------------
Viper, you and I are receding at twice the speed of light from the people (if there are any) living in a certain galaxy which was observed only last year.
--------------------------------------------------------------------------------


Anders said:
--------------------------------------------------------------------------------

ALERT! CONTRADICTION DETECTED! :)
How could we have observed something that is receeding twice the speed of light?
--------------------------------------------------------------------------------

Indeed we have observed something, the quasar (as it was 13 billion years ago), which we believe is currently receding at twice c, and there is no contradiction.

I may have misunderstood Anders comment yet again! But I hope not. Maybe we will hear from him directly :smile:, which would be a definite plus.
 
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  • #15
Hurkyl
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Once something is receding at over the speed of light it can never send a signal that reaches us. Light from that object will never get here.

You sure on that? If, for instance, we start with an object receding special relativistically at a speed of 0.7 c, and then add in a 0.5 c recession due to the expansion of space, the light from the object should have no problem reaching us, although the object is receeding at 1.2 c.
 
  • #16
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Originally posted by Hurkyl
You sure on that? If, for instance, we start with an object receding special relativistically at a speed of 0.7 c, and then add in a 0.5 c recession due to the expansion of space, the light from the object should have no problem reaching us, although the object is receeding at 1.2 c.

This is very interesting, though I have to wonder if Relativity really allows one to make such a distinction about the movement of an object.
 
  • #17
marcus
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Originally posted by Hurkyl
You sure on that? If, for instance, we start with an object receding special relativistically at a speed of 0.7 c, and then add in a 0.5 c recession due to the expansion of space, the light from the object should have no problem reaching us, although the object is receeding at 1.2 c.

just saw your post. no I am not sure what happens in that situation.

I was considering only objects at rest w/rt Hubble flow.
[edit: I must revise what I think even in this case]

But you have posed a much more interesting problem.
there is some expansion and some special relativistic speed.
will consider at soonest possible. why didnt it occur to me? duh.
 
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  • #18
marcus
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Hurkyl

yeah, for sure it reaches us

duh squared
 
  • #19
Hurkyl
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Oh, the expansion of space between us and the theorized current position of that distant quasar is at a 2c rate? How would one compute that?
 
  • #20
marcus
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Originally posted by Hurkyl
Oh, the expansion of space between us and the theorized current position of that distant quasar is at a 2c rate? How would one compute that?

1. they have an idea of being at rest w/rt CMB---same as being
at rest w/rt "hubble flow" the expansion of space.

2. this let's them split the manifold into &Sigma; x R---a space part and a time part

3. the Robertson Walker metric is applied. Also nowadaysow it is assumed (because of CMB observations) that the space part is flat, the RW metric is extremely simple
just has a scale factor a(t) and

ds^2 = -dt^2 + a^2(t)[ euclidean thing]

the scale factor a(t) shows the expansion and it is a solution of
the Friedmann equations, it determines the redshift, and
a t/a is by definition = H0

4. the Hubble law v = H0 D is true at all distances
if D is measured in the present moment ( by the RW metric) and v is the rate of change in that distance.

5. HOW TO CALCULATE starting from z.

there are messy formulas which find the current distance to the quasar from z----these involve the prevailing assumptions
that cosmological const is 0.73 of rho crit. Spatial flatness.
Hubble parameter is 71 km/s per Mpc. I resort to Wrights
online calculator, which has these as default settings but allows
you to put in a different Hubble parameter and different cosmological constant if you want. I use the default settings.

You put in z = 6.4 (which was observed) and it does the messy
part and gives you that the distance to put in the Hubble law is 28 billion LY. this is the present distance also called "comoving" distance---basically just the RW metric in presentday space.

It also tells you how long the light has been traveling---how old the universe was when the light was emitted---and so on.

Then you take the distance D of 28 billion LY and apply the Hubble law and get v = H0 D = 2c

that part is easy because H0 reciprocal is 14 billion Y
so just divide 28 billion LY by 14 billion Y and get 2c.

You might be interested in the formulas and calculating it yourself instead of resorting to the Javascript online calculator.
they are hairy because the redshift was acquired at different epochs of time during different regimes of expansion, what with
cosmological constant egging on the expansion and matter slowing it down and these things being in changing proportions thru time. It is basically a bunch of integrals which people have solved in closed form but there are a lot of terms. I have only rarely tried to compute this myself and am always forgetting terms
So I have come to rely on the calculator.

Would you like me to dig up the formulas though? You may wish to try them. Cant promise that I can lay hands on them but could also look on web.

You also may wish to use the calculator to explore dependence on
parameters so here is the URL for it

http://www.astro.ucla.edu/~wright/CosmoCalc.html
 
  • #21
4. the Hubble law v = H0 D is true at all distances
Do they say that this is instantaneously true? What about C = a/t?
 
  • #22
wimms
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I thought that when recession reaches 1c, distant object is 'switched off' for us. That we can observe only objects that at the time when they emitted their light were receding from us < 1c.

Now you people tell that for expansion, this doesn't count. This is very interesting, because (paraphrasing Hurkyl) IF, for instance, we start with an object receding special relativistically at a speed of 0.0 c, and then add in a infinite c recession due to the expansion of space, the light from the object should have no problem reaching us, although the object is receeding at infinite c.

If objects are stationary in relation to each other relativistically, then they actually can be receeding due to expansion at infinite speed, and we could see them?! wow.

Then what, redshift drops frequency down towards 0Hz, or, static field?

So, emitted photons are not lost, only redshifted and perceived as having reduced energy. As BB is only 14Gyr, before that photons were not free. But, in future, we then should 'see' more and more relict photons building up, coming from areas of space that were far away from us at moment of BB? Jesus, then we should be able to 'watch a movie' of BB occurring just by selecting frequency of relict radiation reaching us.
 
  • #23
marcus
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the zero subscript on H0 refers to the present moment, which is called t = 0.
The Hubble law applies precisely to distances and velocities measured at the present moment in time.

More generally the definition of H is a-dot/a (cant type a with dot over it)
that is time-derivative of scalefactor a(t) divided by scalefactor itself
that is at divided by a.

a-sub-tee is another way to write the time-deriv or rate of change of the scalefactor a

both a and its time-derivative change with time, so the
ratio at / a also changes with time

evaluating this at time zero gives us H0 the present value of the Hubble parameter


Originally posted by wimms
I thought that when recession reaches 1c, distant object is 'switched off' for us. That we can observe only objects that at the time when they emitted their light were receding from us < 1c.

Wimms, I have been thinking about how to answer and have erased my original answer. I just have to think some more about what you say. What I orig. said was too hasty. Sorry. will get back to this as soon as time permits
 
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  • #24
wimms
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Originally posted by marcus
I do not understand this. can't figure out where you got this idea. It sounds quite wrong.
I was confused by Hurkyl's example that you agreed with. I did read it as: if we have object that was relativistically receeding at 0.7c, and it was in point of space that receeded from us at 0.5c due to expansion, back then, then we'll be able to see that light although in total it was receeding at 1.2c. I just exagerated that to infinity, duh. How about 0.99c + 0.99c?

I got impression that if object was receeding at <1c relative to 'intermediate' frame, space, then its light 'escapes' it and further on travels at 'c' towards us. And although space itself expands, in shorter distance expansion speed is <<c, and because any inertial frame is equal, light speed is not 'reduced' over that distance, only redshift occurs.

Anyway, if recessioon of 1c is max we can see, then it must be that in future we'll be seeing limit of U where distant objects are being 'switched off' as we look at them, due to crossing critical limit of 1c recession? Hard to imagine that exact crossover point, one day you receive light at 'c' from it, next day you receive nothing at all.
 
  • #25
Jack
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OK, now for my next stupid question-

If time slows down to a stop as you approach the speed of light then how can light travel?
 
  • #26
Originally posted by Jack
OK, now for my next stupid question-

If time slows down to a stop as you approach the speed of light then how can light travel?

Because GR only applies to objects with rest mass. Photons have no rest mass. They only impart momentum when striking an object; they have have no real mass. By definition a photon cannot have mass. If it did, you could slow it down below its c velocity in a vacuum (a big no-no). Also, there is no time that a photon is accelerating (a GR requirement). It's at c from its creation until you cause it to interact with some medium. That's not to say that you can't modify light in a vacuum. Gravitational blueshifting is an example of wavelength shifting. But it still continues to travel at c.
 
  • #27
marcus
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Originally posted by wimms
I was confused by Hurkyl's example that you agreed with. I did read it as: if we have object that was relativistically receeding at 0.7c, and it was in point of space that receeded from us at 0.5c due to expansion, back then, then we'll be able to see that light although in total it was receeding at 1.2c. I just exagerated that to infinity, duh. How about 0.99c + 0.99c?

.

Hi wimms! I have been away and just got back to board and saw your post, or would have replied sooner.

I tried just now to answer but got off on the wrong foot.
also too many demands on time this morning
had to erase. will get back to this.
 
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  • #28
Hurkyl
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You might be interested in the formulas and calculating it yourself instead of resorting to the Javascript online calculator.

I guess I should have phrased my inquiry better hehe, though I got the answer I wanted: mainly I just wanted to know if the rate of recession is computed between where we are now and where the quasar is "now". (instead of when it was observed) I imagine from your presentation that the actual formula is probably too complicated to be worth studying with my present level of knowledge of GR.
 
  • #29
marcus
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when and if I find anything along those lines I will
send it to you PM

the basic question in my mind is, how does Wright's
online calculator work----how does it get from the
redshift z you put into the current distance to the object
which it give you back.

maybe to your eyes the calculation is not so bad
in anycase no harm in your taking a look and making
your own assessment. I know I am really lazy and would
rather just use the calculator---but I am still curious how
the calculation goes

it may be awhile before I come across it but I will keep
an eye out. also if you happen upon it before I do I'd appreciate a pointer to the information.
 
  • #30
KLscilevothma
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When you say no speeds can be faster than light be sure to specify that it is in the context of Special Relativity. It is true in that context and not true in the universe at large of General Relativity.
Could you please elaborate it by using an example?

I think nothing can travel faster than light. Although quasars have great redshift, but it doesn't mean they are traveling >c.
If we use the classical formula for doppler shift,
v/c = d[lamb]/[lamb]0=z,
we can get something like 4z or 5z, and v>c.
But if we apply the relativistic doppler formula,
v/c = [(z+1)2-1]/[(z+1)2+1],
v/c always smaller than 1 in this case.

Since quasars move in high velocity, relativistic doppler formula should be applied, which implies that quasars can't travel faster than light.
 
  • #31
AndersHermansson
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Originally posted by Hurkyl
You sure on that? If, for instance, we start with an object receding special relativistically at a speed of 0.7 c, and then add in a 0.5 c recession due to the expansion of space, the light from the object should have no problem reaching us, although the object is receeding at 1.2 c.

What's the difference between moving special relativistically and moving due to the expansion of space?
 
  • #32
physicskid
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Since light from very distant objects cannot reach us, there must be a certain horizon in the universe, like the event horizon of a black hole. And this horizon emits thermal radiation, similar to black holes, and they are now observed as small variation in the temperature of the cosmic background.[URL [Broken] a site on hawkig radiation[/U

Is there an equivalent of the sonic boom for light?

Physicskid[zz)]
 
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  • #33
marcus
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Originally posted by AndersHermansson
What's the difference between moving special relativistically and moving due to the expansion of space?

Hello Anders, I just now saw your post. (forgot to check this thread for a while). Hurkyl proposed looking at an object in a part of space receding 0.5 c from us, where the object also has its own individual speed (relative to the CMB) of 0.7c.

The important thing to realize is that the 4D coordinates of SR do not fit space at large scale. So the object as no "special relativity" speed relative to us. It simply is not defined. SR type speeds are defined only for objects at the same point or in the same local coordinate chart.

The metric used in cosmology (socalled RW metric) is defined with reference to observers stationary with respect to the CMB or, as cosmologists say, w/rt the Hubble flow (the expansion of the universe). The RW, or RobertsonWalker, metric can handle the speed Hurkyl mentioned. The combined recession speed, the rate of change in RW distance, is meaningful and equal to 1.2 c.

The RW distance, measured at the present moment, is also called the "comoving distance" or the Hubble-law distance because it is the gauge of distance that works in the well-known Hubble law:

v = H0 D.

The v in the Hubble law is the present rate of change of the RW distance, which is D.

The H0 parameter is, by definition, the present value of

at/a

where a(t) is a numerical valued function of time used to define the RW distance----a distance scale measured from the standpoint of observers at rest with respect to the expansion of space---ie. w/rt the CMB.

a(t) is defined because, in cosmology, all stationary observers everywhere in the universe can have the same time axis---they can have synchronized clocks---essentially one clock for everybody. So there is one scale-function a(t) evolving with time, for everybody.

This is radically different from the picture you get in SR. It is very much a General Relativity picture, by contrast to the local SR picture.

The basic trouble with SR, why it can't be used except locally around a point where there is not much curvature, is that
SR does not allow for space to expand.
The Minkowski 4D coordinates of Special Relativity are rigid and do not let space expand, so these coordinates "fit" the world only in small local patches or local "charts".

this is why the speed of recession of distant objects is not defined in SR coordinates and all that "nothing can exceed the speed of light" business simply does not apply.

Hope this adequately covers the question. the distinction between local SR distance and the real physical distance with the metric is important to make. Please ask further if not clear. I can get you some links to, for instance, Eric Linder's Cosmology Overview (which covers the RW metric) and New Wright's tutorial, which discusses the RW distance and three alternative ideas of distance (angular size, luminosity, light-travel-time). Also other people may want to take a shot at this and you will get several viewpoints.
 
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  • #34
marcus
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Originally posted by physicskid
Since light from very distant objects cannot reach us, there must be a certain horizon in the universe, like the event horizon of a black hole. And this horizon emits thermal radiation, similar to black holes, and they are now observed as small variation in the temperature of the cosmic background.[URL [Broken] a site on hawkig radiation

Is there an equivalent of the sonic boom for light?

Physicskid[zz)]

You have a link here to the Usenet Physics FAQ about cerenkov radiation. The Physics FAQ is a great web resource! It also has something on Hawking radiation. Here is a quote from the Usenet FAQ on Hawking radiation:

(John Baez writes)
"...How does this work? Well, you'll find Hawking radiation explained this way in a lot of "pop-science" treatments:

Virtual particle pairs are constantly being created near the horizon of the black hole, as they are everywhere. Normally, they are created as a particle-antiparticle pair and they quickly annihilate each other. But near the horizon of a black hole, it's possible for one to fall in before the annihilation can happen, in which case the other one escapes as Hawking radiation.
In fact this argument also does not correspond in any clear way to the actual computation. Or at least I've never seen how the standard computation can be transmuted into one involving virtual particles sneaking over the horizon, and in the last talk I was at on this it was emphasized that nobody has ever worked out a "local" description of Hawking radiation in terms of stuff like this happening at the horizon..."

http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/hawking.html

The widely-believed picture of one virtual partner falling in and the other getting away (you give a link to an animation of this idea) may be too simple to explain the actual Hawking radiation.
Do you have a link that, for instance, derives the Hawking temperature using this model?

An interesting question is---why is the Hawking temperature of the black hole given by this formula?

kTHawking = (1/8piM)(hbar c2/G)

In the Hawking formula for the temperature, M is the mass of the black hole, k is Boltzmann's constant, and hbar and c are the usual hbar and c.

The formula, derived in Hawking's orig. 1975 paper, does not seem to come from this picture of "one partner falling in and the other getting away". The formula, however, it what seems to matter because it predicts the temperature of the radiation.

I would like to understand the temperature formula better. Have you found any website that derives it?
 
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  • #35
physicskid
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Hawking radiation[zz)]

Yes! That's definitely a resourceful site for physics fans who think
they know a lot on physics.

To someone far from the black hole, the escaping particles would appear to have been radiated by the black hole. The spectrum of the black hole is exactly what we would expect from a hot body, with a temp. proportional to the gravitational field on the event horizon.

taken from Stephen Hawking's <<The Universe in a Nutshell>>
The temp. of a black hole depends on its mass. So a smaller black hole would be hotter.
 

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