How does a photon's event horizon look like?

In summary, the conversation discusses a Minkowski diagram depicting an event horizon. The diagram shows the path of a particle traveling at the speed of light, which approaches but never reaches the speed of light. The diagram also shows the trajectory of a photon, which would appear as a straight line with slope 1 or -1. If the Minkowski diagram was viewed from the reference frame of the particle traveling at the speed of light, the diagram would collapse and there would be no "t" dimension.
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  • #2
I don't understand the question. Are you perhaps interested in the Rindler horizon experienced by an accelerated observer? (Sorry if this is NOT what you're interested in, but I'm having difficulty in interpreting your question).
 
  • #3
I assumed that the shape of the event cone would depend on the velocity with which point E is traveling along the t axis. Is that not right?
 
  • #4
I don't understand the question either. "Event horizon" of a photon? "Event cone?"

- Warren
 
  • #5
Isn't he showing a Minkowski diagram?
 
  • #6
chroot said:
I don't understand the question either. "Event horizon" of a photon? "Event cone?"

- Warren

DaveC426913 said:
Isn't he showing a Minkowski diagram?
pervect & Warren, thank you for your responses and patience with a non-physicist. Your reply prompted me to take another look at the Wikipedia diagram.

BTW, I noticed that the first link in my OP has become nonfunctional. The new link is: http://upload.wikimedia.org/wikipedia/commons/1/1d/Event-horizon-particle.svgIf you click on it you will see the diagram represented as a spacetime diagram on the Wikipedia page entitled "event horizon."

The caption applicable to this diagram (click on: http://en.wikipedia.org/wiki/Event_horizon) reads: "event E [] is outside the particle [P]'s event horizon."

My questions:
1. Can anyone see the diagram, or is there a broken link?

If you can see the diagram:

2. Which object on the diagram represents P's event horizon?
a. Does the the red curve represent P's event horizon?
b. Do the boundaries of the hourglass-shaped yellow area represent P's event horizon?
c. Does the entirety of the hourglass-shaped yellow area represent P's event horizon?
d. Other: ______ (please specify).

3. What does the entirety of the hourglass-shaped yellow area represent?

4. What do the boundaries of the hourglass-shaped yellow area represent?

5. The Wiki article states "As the particle [P] accelerates, it approaches, but never reaches, the speed of light with respect to its original reference frame. On the spacetime diagram, its path is a hyperbola, which asymptotically approaches a 45 degree line (the path of a light ray)."
5.1. Would this diagram be different if the particle P was traveling at the speed of light?
5.2 How would it be different?
a. Would the red curve be different? How?
b. Do the boundaries of the yellow area be different; how?

Let me know if you need additional information or explanation about any of the questions above and I'll try my best to provide it.
 
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  • #7
Partly due to a lack of responses on the PF, I put the question to Cornell University's "Curious? Ask an Astronomer!" website, and I got an answer in about a week. I decided to post it here for general information and any reactions from the community.

It says (with reference to the Minkoswki diagram on the Wiki page "Event Horizon"):
If event P was the emission of a photon, then the photon's trajectory would appear as a straight line with slope 1 or -1 in the Minkowski diagram (the same as the boundaries of the light cone). The photon could leave event P traveling in the -x direction, in which case it would intersect the light cone, or in the +x direction, in which case it would travel parallel to the right boundary of the light cone for all time.
Admittedly this answers a different question than my original, which I can best state as "If Minskowski was a photon, would there be a diagram? How would it look?" So my original question did evolve under this thread, hopefully for the better.

[Further edit (now this is beginning to look like a blog): Rarely the one to give up, I put my original question to the exceedingly helpful staff of Cornell's Ask An Astronomer website. Their quick response was:
I think I understand your question. The axes of a Minkowski diagram correspond to a particular reference frame. You are curious about what the Minkowski diagram would look like if we were in the frame of an object moving at the speed of light (a photon).

The angle between the time and space axes is 0 degrees if the reference frame is traveling at the speed of light, and so the diagram collapses. So your intuition is good: there would not be a "t" dimension because, in fact, there would be no diagram at all.
Excellent website & great staff, Ezra!]
 
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FAQ: How does a photon's event horizon look like?

What is a photon's event horizon?

A photon's event horizon is a theoretical boundary surrounding the photon where the gravitational pull is so strong that not even light can escape. It is the point of no return for a photon.

How is a photon's event horizon related to black holes?

A photon's event horizon is similar to the event horizon of a black hole. Both are defined as the boundary where the escape velocity exceeds the speed of light. However, a black hole's event horizon is caused by the intense gravitational pull of the singularity, while a photon's event horizon is caused by its own energy.

Can we actually see a photon's event horizon?

No, we cannot see a photon's event horizon as it is a theoretical concept and not a physical object. We can only observe its effects, such as the bending of light near massive objects like black holes.

How does a photon's event horizon affect the path of light?

A photon's event horizon causes light to bend and follow a curved path around massive objects. This is known as gravitational lensing and is one of the key predictions of Einstein's theory of general relativity.

Is a photon's event horizon the same for all photons?

Yes, the event horizon for a photon is the same regardless of its energy or wavelength. This is because the event horizon is determined by the speed of light, which is constant for all photons.

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