Can we determine motion within an enclosed object using energy and mass?

[ande4jo]

The point is that everyone will not agree on the speed of the object relative to Bob (unless I'm not understanding something). Again, it looks like from Alice's perspective the speed of the object coming back towards Bob would be less than 1% c while the speed of the object coming back towards Bob (from his perspective) would be exactly 1% of c.

 

[DrGreg]

The point is that everyone will not agree on the speed of the object relative to Bob (unless I'm not understanding something). Again, it looks like from Alice's perspective the speed of the object coming back towards Bob would be less than 1% c while the speed of the object coming back towards Bob (from his perspective) would be exactly 1% of c.

In relativity, you don't calculate relative speed by subtracting speeds; you use the formula<br /> \frac{u-v}{1-\frac{uv}{c^2}}.<br />If everyone uses the correct formula, they will get the same answer.

 

[Dale]

Thus I still think it's possible the Bob is dead in his frame but not in Alice's frame.

This is not possible. The outcome of any measurement is frame invariant. If after the collision a physician measures Bob's heart rate then all frames must agree on the number that the heart rate monitor displays.

 

[Ibix]

The only speed that matters is the one your device measures, which is speed relative to Bob. The speed someone else measures isn't important - the device doesn't care who is watching.

The point is that it your device will trigger if the speed it measures exceeds a certain value. I can always take my speed measurements and calculate the speed that the device measures by reversing the calculation Janus performed. I will therefore be able to predict the result.

 

[ande4jo]

In relativity, you don't calculate relative speed by subtracting speeds; you use the formula<br /> \frac{u-v}{1-\frac{uv}{c^2}}.<br />If everyone uses the correct formula, they will get the same answer.

Dr Greg
If everyone gets the answer then what is the point of relativity? Your statement implies that there is only one answer and that there is not a need for relativity.

 

[ande4jo]

The only speed that matters is the one your device measures, which is speed relative to Bob. The speed someone else measures isn't important - the device doesn't care who is watching.

The point is that it your device will trigger if the speed it measures exceeds a certain value. I can always take my speed measurements and calculate the speed that the device measures by reversing the calculation Janus performed. I will therefore be able to predict the result.

Ibix
If there is only one frame that is right, then what is the point of relativity? Your statement implies the need to abandon relativity since there is only one valid frame (according to your view) and all others frames are wrong. You might want to reevaluate your logic on relativity if that is how you interpret relativity.

 

[Mentz114]

Ibix
If there is only one frame that is right, then what is the point of relativity? Your statement implies the need to abandon relativity since there is only one valid frame (according to your view) and all others frames are wrong. You might want to reevaluate your logic on relativity if that is how you interpret relativity.

That is completely missing the point. If a time bomb is set to explode when its clock reads 12 say, then everyone watching the bomb will agree that it exploded. If instead of the clock the bomb explodes on some other trigger, then everyone will also agree.

The whole point of relativity is to rule out the kind of ridiculous things you suggest - and it succeeds.

You might want to reevaluate your logic and how you interpret relativity.

 

[PeterDonis]

If everyone gets the answer then what is the point of relativity?

To be able to get the same answer for an observable regardless of what frame you use to calculate it.

If there is only one frame that is right, then what is the point of relativity?

Ibix did not say that only one frame was right. He said that no matter what frame you use, you must get the same answer for an observable. The device triggering is an observable, a physical event; it either happens or it doesn't happen. If it happens, it happens in all frames, and all frames must calculate that it happens. The point of relativity is to show how that works.

 

[Janus]

Ibix
If there is only one frame that is right, then what is the point of relativity? Your statement implies the need to abandon relativity since there is only one valid frame (according to your view) and all others frames are wrong. You might want to reevaluate your logic on relativity if that is how you interpret relativity.

No. it is not just one "valid" frame. All frames are equally valid, it is just that all frames will always agree as to the end result. You have to take everything into account when you transform between frames. For instance let's say the object that hits Bob and kills him penetrates 1/2 the way into his body before it stops. In the frame in which Bob and the ship is moving at 0.99c, Bob is only 1/7 as thick. So if in his own frame the object penetrates 5 in of his 10inch depth, in the other frame the same object has to only penetrate 0.714 into have penetrated halfway into Bob. Then there is the difference in the change in momentum Bob would measure for the object going from 0.1c to 0 over 5 in, vs the other frame where the object goes from 0.988979901 to 0.99c, which represents a much larger momentum change. But does it so over a distance equal to the distance the ship has traveled during the period the object is moving through Bob's body minus 0.714 in. IOW, figuring out what happens according to another frame is much more complicated than just comparing the relative velocity of the object with respect to Bob in that frame. And when you do take everything into account, everyone in every frame will agree that it works out that the object imbedded itself and came to a stop after it had traveled halfway through Bob.

 

[A.T.]

Thus I still think it's possible the Bob is dead in his frame but not in Alice's frame.

Reference frames are not parallel universes or alternate realities, they merely assign different space & time coordinates to the same events.

 

[Ibix]

Let's try a more every day example. Say your car has impact protection that's good for collisions up to 60mph. You are driving down a road doing 35mph according to your speedometer. You see someone coming straight towards you, also doing 35mph according to their speedometer.

According to you, you are stationary and the other car is coming at you at 70mph.

According to the other car, they are stationary and you are coming at them at 70mph.

According to someone crossing the road on foot, you are both coming at him at 35mph.

According to you this is going to hurt you because 70+0>60. According to the other driver this is going to hurt you because 0+70>60. According to the pedestrian (scurrying out of the way) this is going to hurt you because 35+35>60.

What the pedestrian has done, probably without thinking about it, is to transform his speed measurements into the only frame that matters here - yours. Although 35<60, he's realized that that doesn't matter and worked out what it looks like to you.

The only thing the pedestrian has done wrong is to use an incorrect velocity addition formula. The correct one is w=\frac{u+v}{1+uv/c^2}where u and v are 35mph, c is whatever the speed of light is in mph, and w is the speed you want to work out. This mistake really doesn't matter in this example - 1+uv/c^2 is so close to one in this case that you need to worry about the braking effect from bugs hitting your windscreen before you worry about the error from using w\simeq u+v. But in your example this approximate formula (which is what you are using when you simply take the difference between velocities) is a very bad approximation indeed.

Your device can only measure speed relative to Bob, and then act on that. If you are not measuring speed relative to Bob then you need to determine the speed relative to Bob using the correct formula to work out if it will trigger or not. The way you are doing it is to use Janus' relativistic calculation to determine the speeds you would see, then to use a non-relativistic approximation to try to reverse the calculation. That's where the contradiction comes from.

 

[HallsofIvy]

Let's try a more every day example. Say your car has impact protection that's good for collisions up to 60mph.

Once again, you ate (intentionally to point out the error) not saying what frame this speed is in reference to. Is the "impact protection" good for collisions up to 60 mph with, say, a stationary wall?
If so then what is important is the speed of one car relative to another. The relative speeds of the two cars is (in terms of Gallilean relativity) 70 mph to all observers.

[/ quote]You are driving down a road doing 35mph according to your speedometer. You see someone coming straight towards you, also doing 35mph according to their speedometer.

According to you, you are stationary and the other car is coming at you at 70mph.

According to the other car, they are stationary and you are coming at them at 70mph.

According to someone crossing the road on foot, you are both coming at him at 35mph.

According to you this is going to hurt you because 70+0>60. According to the other driver this is going to hurt you because 0+70>60. According to the pedestrian (scurrying out of the way) this is going to hurt you because 35+35>60.

What the pedestrian has done, probably without thinking about it, is to transform his speed measurements into the only frame that matters here - yours. Although 35<60, he's realized that that doesn't matter and worked out what it looks like to you.

The only thing the pedestrian has done wrong is to use an incorrect velocity addition formula. The correct one is w=\frac{u+v}{1+uv/c^2}where u and v are 35mph, c is whatever the speed of light is in mph, and w is the speed you want to work out. This mistake really doesn't matter in this example - 1+uv/c^2 is so close to one in this case that you need to worry about the braking effect from bugs hitting your windscreen before you worry about the error from using w\simeq u+v. But in your example this approximate formula (which is what you are using when you simply take the difference between velocities) is a very bad approximation indeed.

Your device can only measure speed relative to Bob, and then act on that. If you are not measuring speed relative to Bob then you need to determine the speed relative to Bob using the correct formula to work out if it will trigger or not. The way you are doing it is to use Janus' relativistic calculation to determine the speeds you would see, then to use a non-relativistic approximation to try to reverse the calculation. That's where the contradiction comes from.

 

[Ibix]

Once again, you ate
(intentionally to point
out the error) not
saying what frame this
speed is in reference
to. Is the "impact
protection" good for
collisions up to 60 mph
with, say, a stationary
wall?
If so then what is
important is the speed
of one car relative to
another.

That was unintentional sloppiness on my part. I intended that it should be read as you did - thank you.

The relative
speeds of the two cars
is (in terms of Gallilean
erelativity) 70 mph to all
observers.

Indeed. The point (that I hope I got across) is that as you get closer to the speed of light the Galilean approximation gets more and more approximate and the closure rate between two objects in some frame is, in general, nowhere near the velocity of one of the objects in the rest frame of the other.

 

[ande4jo]

Well, if there is a means using relativity to have alternate universes or dead/alive paradoxes then the means to try and bridge relativity and quantum would be somewhere in that answer (since quantum effects and uncertainty principle have this paradox). Also note the best answer that I have seen so far has to do with evaluating more than simply the speed but also the momentum. This is because I envisioned (within my thought experiment) an object that penatrates Bob to kill him and in order for object to penatrates it would need need speed and mass (momentum). I see that relativity would have the object moving toward Bob at a slower rate from Alice's perspective but it would also have the mass increase from her perspective. Thus i assume these 2 persectives cancel each other so that both Alice and Bob would agree on momentum of the event. That said, the are objects that "live" based on time alone and the speed of mesons (relative to Earth observers) indicate that from our perspective mesons live longer when they travel faster than when they are at rest. Last, thank you Mentz114 for telling me that I should check my logic (making me realize the unprofessional nature of my comment to ibix). In my defense I had a couple of beers in me when I posted but that is still excuse so Ibix i am sorry for my ştupid comment.

 

[Nugatory]

Well, if there is a means using relativity to have alternate universes or dead/alive paradoxes then the means to try and bridge relativity and quantum would be somewhere in that answer (since quantum effects and uncertainty principle have this paradox).

Quantum mechanics and special relativity were successfully bridged two generations ago with the discovery of quantum field theory - that's not an unsolved problem. And although quantum mechanics is strangely and wonderfully counterintuitive, it has neither dead/alive paradoxes nor alternate universes (at least as you are using the term).

 

[Nugatory]

Also note the best answer that I have seen so far has to do with evaluating more than simply the speed but also the momentum.

Momentum is every bit as frame-dependent as speed. The two are closely related, and we use momentum more often than speed because it's conserved and it's easier to calculate with in most problems.

 

[ande4jo]

So time, momentum, velocity, mass, and length are all frame dependent? If that's true what is the mechanism to insure that final outcomes of events are always frame independent?
Going back to my previous example of Bob and Alice?

 

[Dale]

So time, momentum, velocity, mass, and length are all frame dependent? If that's true what is the mechanism to insure that final outcomes of events are always frame independent?

Why would there be a mechanism? This is just math. A mechanism implies some sort of cause and effect.

There is one physical scenario. We are describing the same physics in different mathematical ways. If the mathematical descriptions of the final measured outcome do not match the actual measured outcome then they are not valid.

It isn't a matter of a mechanism, it is simply a matter of a set of consistent mathematical descriptions of the same physics.

 

[Nugatory]

So time, momentum, velocity, mass, and length are all frame dependent? If that's true what is the mechanism to insure that final outcomes of events are always frame independent?
Going back to my previous example of Bob and Alice?

Proper time, four-momentum, four-velocity, rest mass, and proper length are all frame-independent and invariant. These are what determine the outcomes of experiments.

The frame-dependent quantities that you mention would more correctly be called coordinate time, three-momentum, relativistic mass, and coordinate length. Although frame-dependent, the mathematical rules that relate these are such that the invariant properties come out the same no matter which reference frame you use to calculate them. As a simple classical example the position of the nose of a fish is frame-dependent, as is the position of the tail of the fish - but the difference between the two positions is the length of the fish and is not frame-dependent (at least classically, and you really do have to understand the classical picture before you can take on special relativity).

 

[ande4jo]

So using my previous example please explain using the math why the object that bounces off of the wall with enough momentum in Bob's frame to kill him but either does not bounce off the wall in Alice's frame or bounces off the wall with less momentum than required to kill Bob in Alice's frame, explain please how these phenomenon all end up with the same outcome.

 

[Nugatory]

So using my previous example please explain using the math why the object that bounces off of the wall with enough momentum in Bob's frame to kill him but either does not bounce off the wall in Alice's frame or bounces off the wall with less momentum than required to kill Bob in Alice's frame, explain please how these phenomenon all end up with the same outcome.

Your triggering device is responding to the magnitude of the four-momentum, which will be the same no matter which frame you use to calculate it.

You can't build a device that triggers off of the value of a frame-dependent quantity such as the three-momentum, for the same reason that you can't build a device that will trigger or not according to whether the person watching (that's "watching"! - not "interacting with"!) the device is moving or at rest relative to the device.

 

[ande4jo]

Thanks.
For those of us that don't understand 3 momentum vs 4 momentum can you give a quick example?

 

[Mentz114]

Thanks.
For those of us that don't understand 3 momentum vs 4 momentum can you give a quick example?

This is not the example you want but you should remember that the relative velocity between the ball and Bob's head is frame invariant.
Everyone sees the ball strike with the same force.

 

[Janus]

Thanks.
For those of us that don't understand 3 momentum vs 4 momentum can you give a quick example?

So using my previous example please explain using the math why the object that bounces off of the wall with enough momentum in Bob's frame to kill him but either does not bounce off the wall in Alice's frame or bounces off the wall with less momentum than required to kill Bob in Alice's frame, explain please how these phenomenon all end up with the same outcome.

But it will have enough momentum to kill Bob is both frames. That's the whole point behind the Lorentz transforms. You start with the fact that anything that happens according to one frame happens according to all frames. Then you consider that c is invariant. The Lorentz transforms are what allows any frame maintain the consistency of both these facts. In other words, in order for Alice to agree with Bob that the object kills him and measure the speed of light relative to herself as being c, she has to measure Bob and his ship as length contracted, time running slow for Bob, and a different clock synchronization than Bob does. The Lorentz transforms are what maintain this consistency of events and the invariance of the speed of light.

Let's take this simple example. IN Bob's frame, we have two objects of 1kg each moving at 0.01c relative to the ship and towards each other until they hit in an non-elastic collision (like two balls of clay) and stick together. The resulting 2kg mass will be motionless with respect to the ship and have 0 momentum as measured by Bob.

Now consider what Alice would conclude. For her the ship (and Bob) are moving at 0.99c. Using the addition of velocities formula it works out that the velocity of one object will be 0.990197c and that of the other object will be 0.989799c. This gives them speeds of 0.000197c and 0.000201c with respect to the ship according to Alice. At first blush it then seems that they could not possibly collide, stick together and have the resulting mass remain motionless with respect to the Ship.

However, this first impression is wrong. Let's work out the whole problem from Alice's frame:

One object with a rest mass of 1kg is moving at 0.989799c, its momentum is \rho = mv\gamma where \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} and m is the rest mass for the object
For ease of calculation, we will use units where c=1, which gives us a numerical value for its momentum of 6.94738.

The other object is moving at 0.990197c and has a momentum of 7.08915( in the same direction). So after the objects collide and stick, the resulting mass will have a momentum of 14.03653. Plugging this and a rest mass value of 2 kg into the momentum formula above gives the resultant velocity of this mass with respect to Alice. This works out to 0.99c, or the same speed as Bob and his ship. In other words, both Bob and Alice agree that according to their own measurements and observations, that the result of the collision is a mass that remains motionless with respect to the ship.

The fact that the two masses had different speeds with respect to Bob as measured from Alice's frame did not, in the end, change the result as far as Alice was concerned. Assuming that it did would lead to a false conclusion. The same type of false conclusion you are making when you assume that an object thrown against a wall and coming back and hitting Bob hard enough to kill Bob in his own frame, would not hit him not hard enough to do so as measured from Alice's frame.

 

[ande4jo]

Thanks a lot for the very detailed explanation. I learned a lot from this thread.