A silly question ?

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Or maybe it isn't and apologies in advance if this isn't the appropriate forum but I was wondering if it is possible to take a photo of something travelling at the speed of light ?

If so, how...if not, why not ?
 

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28,542
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Or maybe it isn't and apologies in advance if this isn't the appropriate forum but I was wondering if it is possible to take a photo of something travelling at the speed of light ?

If so, how...if not, why not ?
No, because nothing massive can travel at the speed of light. You can't take a photo of something that can't exist.

Alternatively you could say yes because light travels at the speed of light and you always take photos of light in some sense, but I don't think that is what you meant.
 
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Ha ha - I was going to say the same thing as DaleSpam - every photo you take is a picture of photons, and they all travel at the speed of light! :tongue:

But no, in practical terms, neither you nor any physical object that you might like to photograph can travel at the speed light, so the answer to your question is, "sorry, no."

On the other hand, there's nothing that says you can't take pictures of objects that move at less than the speed of light, even if they move arbitrarily close to the speed of light. It just takes more and more energy to get them to move that fast, but you can still do it. In fact, that's kind of what particle physicists do at high-energy particle accelerators when detect the by-products of high-energy collisions. In the old days they literally did this with film cameras.
 
tiny-tim
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Welcome to PF!

Hi epithet! Welcome to PF! :smile:

A shadow can move at the speed of light (or faster), and you can take a film of a shadow moving! :smile:

And if a large material object could go faster than light (it can't, of course), then provided it interacted with light, you should be able to see it (either directly, or indirectly by seeing it obscure the background stars beind it) … unless, of course, it's moving directly towards or away from you. :smile:
 
28,542
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A shadow can move at the speed of light (or faster), and you can take a film of a shadow moving! :smile:
Good answer!
 
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Thanks for the welcome :smile:

even for a shadow I was thinking more that by the time the light reaches the shutter the object will have moved on. So would you have to set the shutter speed up to as short a time as possible and point the camera ahead of the object or is that just wishful thinking ?
 
DaveC426913
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Thanks for the welcome :smile:

even for a shadow I was thinking more that by the time the light reaches the shutter the object will have moved on. So would you have to set the shutter speed up to as short a time as possible and point the camera ahead of the object or is that just wishful thinking ?
You're over-thinking it. The light from the object IS where the object is. It's not like if you point your camera ahead of the object it will magically "see" faster than the light can travel.
 
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or, to put it another way, by the time the light reaches you, the object will indeed have moved somewhere, but you don't care - you're still "seeing" the light from where the object was. It's a little like looking at a star that vanished years ago, but whose light we're seeing today. You're seeing history.

That has nothing to do with how fast the object is moving, of course - it's just a matter of how long the light takes to reach you.
 
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I know it seem silly but please indulge me a little more.

I was kinda thinking of light travelling as sound does.

You hear a plane overhead and look up to see where the sound comes from but the plane isn't there, its way ahead so to record the sound at max volume you would point the mic where the sound comes form not where the plane is.

hmmm...so in taking a photo of a star, you're actually taking a photo of the light emanating from it travelling at c and not the star itself as it may have already gone supernova, but thats only because the light is coming straight at you and not moving sideways across the sky like a plane.

If the star were moving across the sky where would you point the camera or would you just not see it or would it look like a comet with a trail ?

I suppose that depends how far away the object is again yeah ?
 
Hootenanny
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Again, I feel that you're over-thinking it, how would you take a picture of the moon?
 
tiny-tim
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You hear a plane overhead and look up to see where the sound comes from but the plane isn't there, its way ahead so to record the sound at max volume you would point the mic where the sound comes form not where the plane is.
Ah … but that's because you use one method (sound waves) for hearing, and another (light waves) for seeing.

Since they travel at different speeds, they come from different directions.

But you and the camera are both using light, travelling at the same speed, and therefore from the same direction! :smile:
 
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If you have an object moving near the speed of light (relative to you), the photons it emits will be highly shifted in frequency due to the relativistic Doppler effect (this is analagous to the shift in sound frequency of passing cars). If the object is moving straight toward you, the photons would be extremely energetic and would probably melt you and your camera (or at least give you a nasty case of cancer). If the object is traveling perpendicular to, or away from you, then the photons would be very non-energetic, leading to a picture so weak that you probably would not see it on film (assuming you're talking about capturing photons in the visible portion of the electromagnetic spectrum).

The relativistic Doppler effect is very straightforward when the object is traveling perpendicular to you at a reasonably large distance. It's identical to the formula related to time dilation / length contraction:

[tex]\nu' = \nu \sqrt{1 - \frac{v^2}{c^2}[/tex]

Here [tex]\nu'[/tex] is the frequency of light emitted when the object is moving, and [tex]\nu[/tex] is the frequency it emits when it is at rest. As the body's velocity increases to the speed of light, [tex]\nu'[/tex] decreases to [tex]0[/tex]. It would become "invisible", even to devices made specifically for capturing and recording ultra-low frequency radio waves. Of course, as mentioned in previous replies, nothing massive can reach the speed of light (relative to you).

This is a fundamental part of Special Relativity, and was introduced in Einstein's first famous paper on the subject of relativity 'On the Electrodynamics of Moving Bodies'.
 
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JesseM
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Here [tex]\nu'[/tex] is the frequency of light emitted when the object is moving, and [tex]\nu[/tex] is the frequency it emits when it is at rest. As the body's velocity increases to the speed of light, [tex]\nu'[/tex] decreases to [tex]0[/tex]. It would become "invisible", even to devices made specifically for capturing and recording ultra-low frequency radio waves. Of course, as mentioned in previous replies, nothing massive can reach the speed of light (relative to you).
I'm not so sure that objects traveling very close to the speed of light would be particularly hard to see--as I said in post #38 in this thread, I don't think the relativistic Doppler formula is meant to apply to reflected light:
The relativistic doppler shift formula is for a source that's emitting radiation at some set frequency in its own rest frame--the reason redshift goes to infinity (i.e. frequency goes to zero) as you approach c is because of time dilation, so if the source is emitting peaks at a frequency of one peak/microsecond in its own frame, in our frame the time between peaks being emitted (not seen) gets longer and longer as the source approaches c, since the time between microseconds on a clock moving along with the object is getting longer in our frame because of time dilation. I'm pretty sure the doppler shift formula doesn't imply that the frequency of light which is reflected off a moving object must go to zero as its speed approaches c (for example, assume the frequency of the incoming light before it hits the object is held constant in our frame, and only the speed of the object is varied).
 
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I'm very sure that the relativistic Doppler effect would apply. I'm not sure what you mean by "reflected light". The object emits a photon, it travels through space, it hits the photoreceptor. I'm not sure where reflection comes into play... Even then, my understanding of mirrors is that they generally transmit photons of the same frequency that impinge upon them.

If this effect did not apply in actuality, then we would not have been able to deduce the rotation curve of a galaxy, or subsequently invent the concept of dark matter.
 
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Again, I feel that you're over-thinking it, how would you take a picture of the moon?
I'd lengthen the exposure time to allow more light to filter into the shutter I suppose. I don't see where your going with this and sorry but all that tech talk just sails right over my head.
 
JesseM
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I'm very sure that the relativistic Doppler effect would apply. I'm not sure what you mean by "reflected light". The object emits a photon, it travels through space, it hits the photoreceptor.
If light hits the object and reflects off it, this isn't really the same as being emitted by the object, is it? It's more like an elastic collision.

Maybe elastic collisions would be the best way to think about this. If we had a train of equally-spaced rubber balls aimed at a large moving wall, and each ball collides elastically with the wall, instantaneously coming back in the opposite direction at the same speed (which we can assume is c since the balls represent photons), will the relativistic Doppler shift formula give the correct answer to the change in spacing between the outgoing train and the ingoing train? I don't think so--for example, if the ingoing train of rubber balls were moving in the +x direction at speed c with a spacing of 10 light-seconds between each successive member of the train (representing the wavelength of a light wave), and the wall is moving in the -x direction at 0.6c, then the time between successive balls hitting the wall will be 6.25 seconds. So if ball #1 hits the wall at t=0 s and position x=10 l.s., with ball #2 at position x=0 l.s. at this moment, then ball #2 will hit the wall at t=6.25 s and position x=6.25 l.s., at which moment ball #1 (moving at c in the -x direction ever since hitting the wall) will be at postion x = 10 - 6.25 = 3.75 l.s. So, the distance between outgoing balls will be 6.25 - 3.75 = 2.5 l.s., or 1/4 the distance between ingoing balls.

In contrast, the relativistic Doppler equation would predict that the wavelength of waves from a source approaching at 0.6c would shrink by a factor of 1/2 relative to the wavelength in the source's frame. Note that in the above I wasn't even considering how things looked in the wall's frame, I was just comparing the wavelength of the ingoing train in my frame to the wavelength of the outgoing train in my frame...because of time dilation, if we see 6.25 s between successive balls hitting the wall, in the wall's frame this is only 0.8*6.25 = 5 seconds between successive hits, so if the wall also sees them moving at c then it must see the wavelength as 5 l.s., which means the wavelength of the reflected train in our frame (2.5 l.s.) is indeed 1/2 of the wavelength of the reflected train in the wall's frame as predicted by the Doppler formula. The point is that although the Doppler formula is still correct for reflection, it isn't telling us what we're interested in here, which is the difference between the wavelength of incoming waves in our frame and the wavelength of the reflected waves in our frame.

However, now that I think of it, in the limit as the wall's frame approaches c, the wavelength of the reflected light does approach zero, i.e. infinite blueshift. If we imagine the same situation as above except with the wall moving in the -x direction at 1c rather than 0.6c, then if ball#1 is hitting the wall at position x=10 l.s. at time t=0, and ball#2 is at position x=0 l.s. at that moment, then ball #2 will hit the wall at x=5 l.s. at time t=5 l.s., and ball #1 will be at exactly the same position at that time.

Still, if we applied the same analysis to a reflecting object neither moving straight towards us or straight away from us, I think the shift in wavelength between the ingoing light and the outgoing light could be by some finite factor, so that in this case we would be able to see the object all right even in the limit as its speed approached c in our frame.
 
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Hootenanny
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I'd lengthen the exposure time to allow more light to filter into the shutter I suppose. I don't see where your going with this and sorry but all that tech talk just sails right over my head.
My point was simply that the moon is moving across the sky and one can easily take pictures of the moon without making any adjustments, as one could do for stars.
 
Ich
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Still, if we applied the same analysis to a reflecting object neither moving straight towards us or straight away from us, I think the shift in wavelength between the ingoing light and the outgoing light could be by some finite factor, so that in this case we would be able to see the object all right even in the limit as its speed approached c in our frame.
You just square the formula, this will make things always worse, never better.
 
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My point was simply that the moon is moving across the sky and one can easily take pictures of the moon without making any adjustments, as one could do for stars.
I don't know if this is epithet's concern, but there is a problem with taking pictures of the moon, and especially the stars, which is that they move during the interval that your shutter is open. For the moon this is not typically a problem, since the shutter need not be open for very long, but with stars it causes the "streaks" that you see in many photographs of the night sky.

In any case, to photograph something moving near the speed of light, you'd have to have an extremely fast shutter speed, or you'd end up with a smeared image.

Otherwise, say if you imagined you had a camera that could form an image from the light that strikes its focal plane at one discrete instant, then I don't see why you'd have any trouble photographing something moving at the speed of light, or even faster, like a shadow that JesseM suggested. You'd simply capture an image formed from the light that came at one instant in time from the "something" (I won't call it an object, since shadow aren't objects). In other words, with this infinitely fast shutter speed, you'd "freeze" any object, no matter how fast, and thus you couldn't tell in the resulting photograph how fast it was moving.
 
russ_watters
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In any case, to photograph something moving near the speed of light, you'd have to have an extremely fast shutter speed, or you'd end up with a smeared image.
Or a distant object and/or a tracking mount. The linear speed of the object doesn't automatically imply any problems in photographing it, only the angular speed across your field of view.
 
JesseM
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You just square the formula, this will make things always worse, never better.
Square what formula, to get the answer to what question? In the above I was calculating the difference between the frequency of an ingoing wave to an outgoing wave after being reflected by an object coming straight at us, and found the answer was 1/4, which is the square of the answer to the question about the difference in frequency between a wave emitted by a source coming straight at us and the frequency of the wave we receive, as predicted by the Doppler formula. But then at the end of the post I asked a separate question about how the frequency would change if the object was not coming straight at us, but instead had some velocity in the direction perpendicular to the direction of the ingoing light. It seems to me that even in the limit as the object's velocity approaches c, in this case the shift in frequency would not be infinite--in fact, if the object were moving exactly perpendicular to the direction of the incoming light, I don't think there would be any shift in the frequency of the incoming light vs. the frequency of the outgoing reflected light in our frame.
 
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I don't think the relativistic Doppler formula is meant to apply to reflected light:
I think you are correct. The relativistic Doppler is simply the normal Doppler with time dilation in. If you are using reflected light then I don't think that the time dilation applies so you would use the classical Doppler formula.
 
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I'm still not sure why you're referring to reflected light. The OP is clearly asking about taking a direct picture of an object.

Can you please explain to me why imaging of galactic constituents confirms the relativistic Doppler shift, and yet this other object would not?

This insinuation that the photons will automatically revert back to the "rest frequency" upon reflection implies that the imaging of galactic constituents would not show any red/blueshift. The point of a dish receiver is that the parabolic shape reflects and focuses incident photons toward a central receiver. Observing the red/blueshift using such a device would be impossible if this "reversion to rest frequency" implication held. Likewise for detectable binary stars of high velocity such as the one in PSR 1913+16.

P.S. I am being absolutely serious about seeing this from your point of view. I find this discussion to be extremely enjoyable, and quite refreshing compared to some "discussions" that I've had with other members of these fora in the past.
 
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JesseM
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I'm still not sure why you're referring to reflected light. The OP is clearly asking about taking a direct picture of an object.
And if you try to take a picture of an object that doesn't emit light on its own--like a planet or asteroid moving at close to the speed of light rather than a star--wouldn't you be looking at reflected light? These objects wouldn't emit a significant amount of photons in an otherwise dark universe (just black body radiation), when we look at them we're seeing the light they reflect from the Sun.
shalayka said:
Can you please explain to me why imaging of galactic constituents confirms the relativistic Doppler shift, and yet this other object would not?
Because these "constintuents" are things like stars that emit their own light rather than just reflecting it.
shalayka said:
This insinuation that the photons will automatically revert back to the "rest frequency" upon reflection
How did you get that out of anything I wrote? I calculated that if an object is moving towards you at 0.8c, the frequency of the reflected light is higher than the frequency of impinging light (both as measured in your frame) by a factor of 4. My only assumption was that each peak of the wave moves towards the object at c and then immediately bounces back at c after hitting it, as measured in our frame.
 

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