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

[Pierre007080]

I am not sure that I am using the correct term of "equivalence" for this question, but here goes. If I were traveling at the speed of light and tried to shine a flashlight forward, would it send out a light beam? Would the same thing not happen if I were standing on the event horison of a Black Hole and trying to shine a flashlight away form the Black Hole?

 

[nicksauce]

The premise of your question violates the laws of physics, so you're probably not going to get any terribly insightful answers.

 

[Pierre007080]

Thanks for the response. Perhaps I should say that I am traveling NEAR the speed of light or NEAR the event horison of a black hole with my flashlight. Does that make more sense?

 

[JesseM]

Thanks for the response. Perhaps I should say that I am traveling NEAR the speed of light or NEAR the event horison of a black hole with my flashlight. Does that make more sense?

As long as you are in freefall, locally you will always observe light to leave your flashlight at exactly c (locally meaning you are making measurements in a small region of space and time where the effects of spacetime curvature aren't noticeable). Light always moves at c in every inertial reference frame in special relativity ('inertial' meaning a frame that isn't accelerating), and the fact that the local observations of freefalling observers in general relativity match up with those of inertial observers in special relativity is known as the http://www.aei.mpg.de/einsteinOnline/en/spotlights/equivalence_principle/index.html . This would even be true for a freefalling observer at the instant he is crossing the black hole's event horizon (the horizon itself is locally seen to be moving outward at a speed of c for such an observer, by the way).

Also, keep in mind that there is no absolute notion of speed in relativity so "traveling near the speed of light" has no absolute meaning. If I am traveling away from you at near the speed of light in the frame where you are at rest, then there is another frame where I am at rest and it is you who is traveling at near the speed of light away from me, and both frames are considered equally valid ways of looking at the situation, there is no physical truth of the matter about who is "really" moving faster.

 

[Pierre007080]

Hi JesseM. Thanks for your reply. Is there no sort of theoretical "bird's eye" view that would place the observer in a position to be able to observe light being emitted from an object in the same direction that the object is moving at nearly the speed of light. Something like Einstein's train and platform. Say that the train was moving near the speed of light and someone directed a flashlight in the direction of travel?

 

[JesseM]

Hi JesseM. Thanks for your reply. Is there no sort of theoretical "bird's eye" view that would place the observer in a position to be able to observe light being emitted from an object in the same direction that the object is moving at nearly the speed of light. Something like Einstein's train and platform. Say that the train was moving near the speed of light and someone directed a flashlight in the direction of travel?

As I said, there is no objective truth about whether anything is "moving at nearly the speed of light" or at rest; there is one frame where the tracks are at rest and the train is moving fast, and another where the train is at rest and the tracks are moving fast. All the laws of physics appear to work exactly the same in both frames. If someone shines a light forward in the direction of the front of the train, then both frames will judge that the light moves at exactly c. Keep in mind that each frame uses rulers and clocks at rest in that frame to measure speed (speed is just distance/time), and each frame judges rulers and clocks in the other frame to be very distorted, with moving rulers shrinking (length contraction) and moving clocks slowing down (time dilation), not to mention that a pair of clocks which are synchronized in one frame will be out-of-sync in the other (the relativity of simultaneity). If you're interested I can give you a numerical example showing how these factors come together to ensure that both frames measure the same beam of light to move at c when they use their own ruler and clocks to measure it.

 

[member 11137]

Hi JesseM. Thanks for your reply. Is there no sort of theoretical "bird's eye" view that would place the observer in a position to be able to observe light being emitted from an object in the same direction that the object is moving at nearly the speed of light. Something like Einstein's train and platform. Say that the train was moving near the speed of light and someone directed a flashlight in the direction of travel?

Do you know that speeds v and v' (same direction) do not simply add; in extenso you don't get : v + v'!

Questions: Is your question trying to explore the idea that any particle at c speed is a BH? Do you indirectly mean that any particle "strahlt" when its speed is slowdown?

 

[Pierre007080]

Thanks again. I think I am starting to understand what I am trying to ask... if that makes sense. The person on the fast moving train will not realize that his rulers and clocks are distorted, but the observer on the platform would surely see the rulers and clocks distorted according to Lorentz's equations. The question thus becomes: would the platform observer see the light emitting forward like the guy on the train or would he see a distorted wave (if he could see it) with a shortened wavelength and a slower propagation (Time dilation)??

 

[Pierre007080]

Do you know that speeds v and v' (same direction) do not simply add; in extenso you don't get : v + v'!

Questions: Is your question trying to explore the idea that any particle at c speed is a BH? Do you indirectly mean that any particle "strahlt" when its speed is slowdown?

Yes to your first question. I do realize that the addition of velocities is not straight forward addition.
I do not understand the term "strahlt". Please help me here.

 

[JesseM]

Thanks again. I think I am starting to understand what I am trying to ask... if that makes sense. The person on the fast moving train will not realize that his rulers and clocks are distorted, but the observer on the platform would surely see the rulers and clocks distorted according to Lorentz's equations.

You're still talking in absolute terms. In the frame of the platform, the rulers and clocks of the guy on the train are distorted...but in the frame of the train, the rulers and clocks of the guy on the platform are distorted! There is no physical truth about whose rulers and clocks are "correct" and whose are distorted.

The question thus becomes: would the platform observer see the light emitting forward like the guy on the train or would he see a distorted wave (if he could see it) with a shortened wavelength and a slower propagation (Time dilation)??

The wavelength would be different in the two frames due to the Doppler effect, but as I said, both frames will measure the speed of the light wave to be exactly c--it's one of the two basic postulates of SR that all inertial frames measure light to move at c (the other postulate is that the laws of physics work the same in all inertial frames).

 

[Pierre007080]

Thanks for your patience, guys! I think I understand the addition of velocities of relativistic speeds never exceeding the speed of light and can relate the shortened wavelength and time dilation brought about by the Doppler effect. Special Relativity features.
This brings me to the second part of my original question relating to General Relativity. I gather that space is warped NEAR any massive object (such as near a black hole). Does that space warp also bring about shorter rulers and time dilation similar (equivalent?)to the situation on the fast train?

 

[Dale]

If you have a very large black hole, such that the tidal forces at the event horizon are approximately zero, then that is equivalent to a Rindler accelerating observer in flat spacetime. Not an inertial observer at any speed.

 

[Pierre007080]

If you have a very large black hole, such that the tidal forces at the event horizon are approximately zero, then that is equivalent to a Rindler accelerating observer in flat spacetime. Not an inertial observer at any speed.

Wow! Do you guys know how to speak English? Please explain: "Tidal forces", "Rindler accelerating observer" and also explain why you are referring to "flat spacetime" when we are taliking about space being warped near a massive body?

 

[Dale]

Wow! Do you guys know how to speak English? Please explain: "Tidal forces", "Rindler accelerating observer"

Pierre, if you don't understand these words then do a search. If you want to learn then you really need to put in a certain minimal amount of effort on your own. There are lots of good resources on the internet for each of these topics, as well as lots of threads on this forum for each of them.

also explain why you are referring to "flat spacetime" when we are taliking about space being warped near a massive body?

The various equivalence principles are all about how certain limiting cases of a curved spacetime are the same as standard Minkowski flat spacetime. Since that is the essence of your OP I am a little surprised by this comment of yours.

 

[JesseM]

Wow! Do you guys know how to speak English? Please explain: "Tidal forces", "Rindler accelerating observer" and also explain why you are referring to "flat spacetime" when we are taliking about space being warped near a massive body?

Please read http://www.aei.mpg.de/einsteinOnline/en/spotlights/equivalence_principle/index.html which I linked to in my first post on this thread. It will explain about tidal forces, and about how situations in small regions of curved spacetime are equivalent to certain situations in flat spacetime. Once you have an idea of that stuff, it will be easier to tackle the question of what a "Rindler accelerating observer" in flat spacetime is.

 

[Pierre007080]

Thank you JesseM. I have re-read the equivalence and the tidal flow stuff. I tried to read again about Minovsky space and Rindler co-ordinates but I'm afraid that it goes beyond my ability to visualise these concepts. Respectfully, I do not want to accept that physics belongs only to mathematicians and would hope that the complicated stuff could be simplified for dummies like me and explained without the complicated mathematics. My simplistic understanding of general relativity is that space is more dense around massive objects. This obviously decreases with distance form the mass. Surely these concentric iso-density rings of decreasing density of space in essence relate to rod length (perpendicular to the rings) and time dilation from ring to ring... I'm not asking for a quantitative answer, but merely a practical way to view this concept before taking the next step. I hope this does not come across as offensive in any way because it is most certainly not intended.

 

[JesseM]

My simplistic understanding of general relativity is that space is more dense around massive objects. This obviously decreases with distance form the mass. Surely these concentric iso-density rings of decreasing density of space in essence relate to rod length (perpendicular to the rings) and time dilation from ring to ring... I'm not asking for a quantitative answer, but merely a practical way to view this concept before taking the next step. I hope this does not come across as offensive in any way because it is most certainly not intended.

I haven't seen any explanation of general relativity that talks about the "density" of space, I don't know what that would mean...usually, simplified explanations talk about the curvature of space near a massive object, like a trampoline with a bowling ball on it. But the type of intuitive argument DaleSpam was referring to was based on the equivalence principle, which doesn't require any math to understand...whatever would be seen by an observer in a small room at a fixed radius from a black hole, that should be equivalent to the observerations of an observer in a room that's being accelerated in flat spacetime, who would feel G-forces as a consequence of acceleration, see light beams curve towards the floor (see the third animation on http://astronomy.colorado.edu/astr2030/astr2030images.html , which shows how a light beam traveling in a straight line will seem to curve towards the floor in a room that's accelerating upwards), and all the other phenomena associated with gravity (except tidal forces, since we are considering a room small enough, and a period of time brief enough, that they aren't detectable for the observer in the gravitational field).

 

[Pierre007080]

Hi JesseM. Thanks for your reply. I think that I am following now. There are two questions that I would like to ask:
• The beam traveling horisontally across the room bends down from the gravity … if it was traveling vertically upwards (directly away from the centre of the large mass) would the wavelength be shortened compared to if the room was further away form the mass? In fact, if we were at (or near) the event horison would the wavelength not be almost infinitely shortened? The observer in the room would not realize it because his ruler would also be shortened? (This is what confused me into thinking that the situation was equivalent to someone traveling at speed trying to measure his shortened wavelength in the direction of travel with his shortened ruler)
• The mass on the trampoline analogy. The implication as I understand it is that a rolling ball would go down the gradient along a line that depended how it approached the depression and at what speed. We are used to seeing geographical maps that show isobars (pressure or height etc). These do not primarily show the path water etc would flow but join points of equal potential. If the trampoline analogy was viewed as layers of iso-potential (gravity or density?) in 3D would the picture not be one of concentric iso-potential spheres around the massive object?
Kind regards.

 

[Dale]

Respectfully, I do not want to accept that physics belongs only to mathematicians and would hope that the complicated stuff could be simplified for dummies like me and explained without the complicated mathematics.

This kind of comment always bothers me. Math is not an optional accessory to physics, if you really want to learn physics you cannot avoid math.

However, the key point is that the statement "I am traveling near the speed of light" implies an inertial observer while "I am hovering near the event horizon" implies an accelerated observer. The two are not equivalent. On the other hand, the statement "I am accelerating in free space" implies an accelerated observer which can be equivalent to "I am hovering near the event horizon".

 

[Pierre007080]

Hi DaleSpam, Thanks for your response. Perhaps my understanding would be helped along if we dropped the black hole and event horison concept and replaced it with a large mass just short of a black hole mass. Then we could concieve standing on the surface of the mass shining a flashlight directly away from the centre of the mass. In this way we could overcome the need for "hovering" or accelerating. Does this make the person an inertial observer?

 

[JesseM]

Hi DaleSpam, Thanks for your response. Perhaps my understanding would be helped along if we dropped the black hole and event horison concept and replaced it with a large mass just short of a black hole mass. Then we could concieve standing on the surface of the mass shining a flashlight directly away from the centre of the mass. In this way we could overcome the need for "hovering" or accelerating. Does this make the person an inertial observer?

Any observer who is not in freefall because they have some non-gravitational force acting on them is non-inertial in general relativity. If you're standing on a planet, it's because electromagnetic forces from the ground under your feet are pushing you up to keep you from falling towards the center as you would "naturally" if no non-gravitational forces acted on you.

 

[Dale]

In this way we could overcome the need for "hovering" or accelerating. Does this make the person an inertial observer?

No, such an observer is both non-inertial and accelerating (proper acceleration). Do you know what an accelerometer is (the 6 degree of freedom kind)? An accelerometer attached to such an observer reads an acceleration of 1 g upwards.

 

[Pierre007080]

Hi Guys, thanks for your clear and definite answers. I feel much more comfortable about the observer's position being an accelerated one. I would appreciate it if you could return to my two questions about:
1. the wavelength shortening of a "vertical" beam?
2. the trampoline analagy?
Regards.

 

[A.T.]

2. the trampoline analagy?

What you describe is the wrong picture. It ignores the time-dimension. See this thread:
https://www.physicsforums.com/showthread.php?p=2046692

 

[Dale]

1. the wavelength shortening of a "vertical" beam?

I think you are asking about gravitational redshift, but I cannot tell exactly what your question is.

 

[Pierre007080]

Hi JesseM. Thanks for your reply. I think that I am following now. There are two questions that I would like to ask:
• The beam traveling horisontally across the room bends down from the gravity … if it was traveling vertically upwards (directly away from the centre of the large mass) would the wavelength be shortened compared to if the room was further away form the mass? In fact, if we were at (or near) the event horison would the wavelength not be almost infinitely shortened? The observer in the room would not realize it because his ruler would also be shortened? (This is what confused me into thinking that the situation was equivalent to someone traveling at speed trying to measure his shortened wavelength in the direction of travel with his shortened ruler)
• The mass on the trampoline analogy. The implication as I understand it is that a rolling ball would go down the gradient along a line that depended how it approached the depression and at what speed. We are used to seeing geographical maps that show isobars (pressure or height etc). These do not primarily show the path water etc would flow but join points of equal potential. If the trampoline analogy was viewed as layers of iso-potential (gravity or density?) in 3D would the picture not be one of concentric iso-potential spheres around the massive object?
Kind regards.

Hi there Guys. I have quoted my two questions that I asked to return to. Maybe my first does refer to gravitational redshift, but surely the shortened wavelength due to gravity(?this was my question) would rather be a blueshift?

I read the comments about the trampoline analogy and just read about how it depicts time and GR ... sorry guys, I must be blinder than I thought. I need more guidance than that please.

 

[Dale]

Maybe my first does refer to gravitational redshift, but surely the shortened wavelength due to gravity(?this was my question) would rather be a blueshift?

Red or blue depends on if the light is going up or down. If the transmitter is lower than the receiver then there is a redshift. If the transmitter is higher than the receiver then there is a blueshift.

 

[Pierre007080]

I think I understand that, but I thought that the redshift was due to the time dilation ... what I am trying to get at is that wavelength is SHORTENED due to gravity. I understood that the redshift was made up of a SHORTENED wavelength, but because time dilation escalates more quickly than the rod (or in this case wave) shortening as we get nearer to the mass, the overall effect would be one of redshift?

 

[A.T.]

I read the comments about the trampoline analogy and just read about how it depicts time and GR ... sorry guys, I must be blinder than I thought. I need more guidance than that please.

The point is, that the trampoline analogy doesn't depicts how gravity works in GR. Follow the links in this post for better analogies:
https://www.physicsforums.com/showthread.php?p=2046692

 

[Pierre007080]

I agree with some of the guys that the trampoline analogy says very little except to show that space can warp. It gives no three dimensional depiction and certainly no time reference.

 

[Pierre007080]

DaleSpam and JesseM it seems as though you have abandoned me?? There seems to be consensus that gravitational redshift would be caused by time dilation, but the question remains: is the wavelength shortened by gravity if the wave is leaving a massive object along ta radial?

 

[JesseM]

DaleSpam and JesseM it seems as though you have abandoned me?? There seems to be consensus that gravitational redshift would be caused by time dilation, but the question remains: is the wavelength shortened by gravity if the wave is leaving a massive object along ta radial?

No, "gravitational redshift" deals precisely with the issue of light emitted from a point closer to the source of gravity to a point farther from the source, and it says that the wavelength is lengthened, not shortened.

 

[Pierre007080]

Red or blue depends on if the light is going up or down. If the transmitter is lower than the receiver then there is a redshift. If the transmitter is higher than the receiver then there is a blueshift.

This comment of yours has been worrying me. If redshift is caused by time dilation, then the direction should surely not matter?

 

[JesseM]

This comment of yours has been worrying me. If redshift is caused by time dilation, then the direction should surely not matter?

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.

 

[Dale]

DaleSpam and JesseM it seems as though you have abandoned me??

There seems to be consensus that gravitational redshift would be caused by time dilation, but the question remains: is the wavelength shortened by gravity if the wave is leaving a massive object along ta radial?

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.