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Relativistic-speed cameras [closed]
I am asking on behalf of someone else. Here is their question:
Two cameras, each recording video at 500 trillion fps, are speeding toward each other at $0.6c$ , on a direct collision course. When the cameras collide, they eject their data drives so it can be collected and reviewed.
However, this means that one camera will never capture the other camera, as it is moving at $1.2c$ relative to the camera. So would this be true, and if not, why?
My note: I am thinking that each camera will be able to see the other camera, due to time dilation and other relativistic effects.
- homework-and-exercises
- special-relativity
- speed-of-light
- time-dilation
3 Answers 3
in special relativity, you have to use the relativistic velocity addition formula , to transfer to the observed velocity in the frame of one of the cameras:
$$v = \frac{v_{1} + v_{2}}{1 + \frac{v_{1}v_{2}}{c^{2}}}$$
so, your camera would observe the other camera moving at:
$$v = \frac{0.6c + 0.6c}{1 + 0.36} = 0.882 c$$
$$0.6c + 0.6c \ne 1.2c$$
It's the classical point of view that speeds simply add. For common speeds in our lives it is a very good approximation, but for such high speeds is necessary to use the exact (“relativistic”) formula
$$\frac{u + v}{\color{red}{1 + \frac{uv}{c^2}}}.$$
As you may see, the denominator $\color{red}{1 + \frac{uv}{c^2}}$ is for low speeds $u, v$ almost $1$ because $c^2$ is an incredibly large number, so the result $(u+v)$ is a very good approximation.
But in your case of very high speeds, $u = v = 0.6c,$ the denominator $\color{red}{1 + \frac{uv}{c^2}}$ makes a difference, because its value is $\color{red}{1.36}$ ; instead of their sum you will obtain
$$\frac{0.6c + 0.6c}{1 + \frac{0.6c\cdot 0.6c}{c^2}} = \frac{1.2c}{1 + \frac{0.36c^2}{c^2}} = \frac{1.2c}{1 + 0.36} = \frac{1.2c}{\color{red}{1.36}} \approx 0.88c $$
- 6 $\begingroup$ now we gotta do colored formulas? $\endgroup$ – JEB Commented Mar 26, 2021 at 2:39
While @JerrySchirmer and @MarianD gave good answers, here's a spacetime diagram that explains why it is $(15/17)c\approx 0.88c$ .
(The rotated graph paper helps us visualize the ticks along observer worldlines [timelines] and spacelines. The "light-clock diamonds" (traced out by rays in a light-clock) have equal areas, as required by the Lorentz transformation.)
In the Lab Frame (along LOZ), before the collision at event Z
- the frontward-bound camera (along BZ) has velocity $\displaystyle\frac{BO}{OZ}c=\frac{6}{10}c$ , using distant-event B and local-event O that LOZ says are simultaneous. Thinking vectorially, $\vec {BZ}=\vec{BO} + \vec{OZ}$ .
- the rearward-bound camera (along AZ) has velocity $\displaystyle\frac{AO}{OZ}c=\frac{-6}{10}c$ , using distant-event A and local event O that LOZ says are simultaneous. Thinking vectorially, $\vec {AZ}=\vec{AO} + \vec{OZ}$ .
In the rearward-bound camera frame (along AZ), before the collision at event Z
- the lab frame (along LOZ) has velocity $\displaystyle\frac{PN}{NZ}c=\frac{6}{10}c$ , using distant-event P and local-event N that AZ says are simultaneous. Thinking vectorially, $\vec {PZ}=\vec{PN} + \vec{NZ}$ .
- the frontward -bound camera frame (along BZ) has velocity $\displaystyle\frac{BM}{MZ}c=\frac{15}{17}c\approx 0.88c$ , using distant-event B and local-event M that AZ says are simultaneous. Thinking vectorially, $\vec {BZ}=\vec{BM} + \vec{MZ}$ .
The key is that velocity-measurements (say, of BZ) by an observer (like AZ) requires a purely-spatial "rise" (like BM) [along simultaneous events] and a purely-temporal "run" (like MZ) [along same-place events] according to that observer.
You can use the rotated graph paper gridlines to draw in what each observer “sees” (by receiving light signals on their worldlines that were broadcast by other observers... that is, “read the clock readings sent from distant worldlines”). The visual comparison of the received distant-clock timestamps with the local-clock timestamps involves the Doppler factor $k=\sqrt{\frac{1+v}{1-v}}$ . (For $v=(3/5)c$ , we have $k=2$ , which is why those diamonds are stretched by a factor of 2 in the forward direction and shrunk by 2 in the other. For $v=(15/17)c$ , we have $k=4$ .) [ $k$ is an eigenvalue of the Lorentz boost transformation.]
Incidentally, using rapidity, $(15/17)=\tanh(\mbox{arctanh}(\frac{3}{5})+\mbox{arctanh}(\frac{3}{5}))$ and $2=\exp( \mbox{arctanh}(\frac{3}{5}))$ and $4=\exp( \mbox{arctanh}(\frac{3}{5}) + \mbox{arctanh}(\frac{3}{5}) )$ .
When the velocities yield rational values for $k$ , the arithmetic involves simple fractions since rational $k$ 's are associated with Pythagorean triples. That's why $v=(3/5)c$ is arithmetically simpler than $v=(1/2)c$ .
Not the answer you're looking for? Browse other questions tagged homework-and-exercises special-relativity speed-of-light time-dilation camera or ask your own question .
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Does the Thin Lens Formula Apply to Pinhole Cameras?
- Thread starter cybernerd
- Start date Jul 21, 2008
- Tags Camera
- Jul 21, 2008
Homework Statement
Homework equations, the attempt at a solution.
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cybernerd said: Homework Statement It's not a specific problem, persay... I'm working on a science lab and I'm supposed to be deriving the height of an image created by a pinhole camera. Since I'm in a condensed Physics course, we are expected to figure out formulas on our own. I keep finding this formula: (height of image) / (height of object) = ( - distance of image) / (distance of object) And since Di = Do, it should work. The problem is, that formula is associated with curved mirror questions...a pinhole camera is something else. Will that formula work? It's the only one I can find to determine the height of image from distance of object and height of object. I really need help with this, I'm completely stuck and I have no idea what's right and what isn't.
- Jul 28, 2008
FAQ: Does the Thin Lens Formula Apply to Pinhole Cameras?
1. what is a pinhole camera.
A pinhole camera is a simple type of camera that works based on the principle of light passing through a small hole or aperture and projecting an inverted image onto a screen or film.
2. How does a pinhole camera work?
A pinhole camera works by allowing a small amount of light to pass through a tiny hole in a light-proof box. This light then projects an inverted image onto the back of the box, which can be captured on film or a digital sensor.
3. What are the equations used in a pinhole camera?
The two main equations used in a pinhole camera are the pinhole camera formula, which calculates the distance from the pinhole to the image plane, and the angular magnification formula, which calculates the size of the image compared to the size of the object.
4. How are the pinhole camera equations derived?
The pinhole camera formula is derived using basic geometry principles, such as similar triangles. The angular magnification formula is derived using the properties of similar triangles and basic trigonometry.
5. What factors can affect the accuracy of the pinhole camera equations?
The accuracy of the pinhole camera equations can be affected by factors such as the size of the pinhole, the distance between the pinhole and the image plane, and the wavelength of light being used. Other factors, such as the quality of the film or sensor used, can also impact the accuracy of the equations.
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2.000 Homework # 1: Disposable Camera 4. UHTW-IV: PHYSICS [ 2pts each ] Explain the physics that can be used to model the machine. No credit if you don't relate the form and/or flow(s) to the physics. Answers are 2 pts each. ... 2.000 Homework # 1: Disposable Camera 5. UHTW-V: FABRICATION [ 2 pts each, 6 pts max ]
Flash in cameras is a burst of bright light that is used to illuminate a scene. It typically consists of a small, built-in light or an external flash unit. When the shutter button is pressed, the flash emits a burst of light that helps to illuminate the subject, especially in low light conditions. This allows for a brighter and more evenly lit ...
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Homework-like questions and check-my-work questions are considered off-topic here, particularly when asking about specific computations instead of underlying physics concepts. Homework questions can be on-topic when they are useful to a broader audience. Note that answers to homework questions with complete solutions may be deleted!
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Created by. Petersonic Science. This bundle contains all the labs we use during the "images in mirrors" section of geometric optics. Answers are included! 1. Using a Pinhole Camera2. Pinhole Camera Worksheet3. Laws of Reflection in a Plane Mirror + Worksheet4. Using Light Rays to Locate Images in a Plane Mirror5.
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In order to accurately determine the height of an image created by a pinhole camera, we need to use the pinhole camera equation: 1/s + 1/s' = 1/f. Where: s = distance of object from pinhole. s' = distance of image from pinhole. f = focal length of pinhole camera.