Alice is running very fast holding a long pole. The pole is held parallel to the direction in which she is running. She’s running into a barn with an open door. When the pole is stationary relative to the barn, it does not fit inside it completely. A tiny bit sticks out, preventing the door from being shut. However, since the pole is now in motion, Bob, who is standing by the barn door, sees its length contracted, allowing it to fit completely inside the barn. This means that Bob can wait for Alice to enter the barn and then shut the door. Let’s say he shuts the door as soon as the leading edge of the pole makes contact with the barn wall opposite the door. But from Alice’s perspective, it is the barn’s length that has contracted, not the pole’s. From her perspective, how do you account for the fact that she is able to run the pole completely into the barn so that Bob can shut the barn door?
The usual response to the twin paradox is that the twins’ situation is not symmetrical because the one who leaves Earth must undergo non-inertial motion in order to turn around and return. However, it is possible to reproduce the twin paradox without either twin accelerating, and without moving to curved spacetimes. Can you think of how one might construct this situation?
We all know that according to special relativity it is impossible for a massive object to accelerate until it catches up with a photon. However, it is possible for a massive object to accelerate so that, in its rest frame, its distance from a photon always remains the same. What would the word-line of such an object look like?
We have two spaceships A and B, initially flying at the same uniform velocity. A is flying behind B, and there is a taut string connecting the nose of A to the tail of B. The spaceships both accelerate so that an observer who starts out at rest relative to them will measure the distance between them as unchanging throughout the acceleration. When the ships stop accelerating they are moving relative to the observer. This means the observer should see the string as Lorentz contracted, but, by hypothesis, the distance between the spaceships unchanged. Does this mean the string breaks? Describe what happens from the perspective of the space-ships.
Sbe Nyvpr, gur cbyr’f raq uvggvat gur onpx jnyy naq Obo pybfvat gur qbbe jvyy abg or fvzhygnarbhf. Jura gur cbyr uvgf gur onpx raq bs gur jnyy, fbzr bs vg jvyy fgvyy or fgvpxvat bhg bs gur onea. Ohg gung ovg jvyy pbagvahr zbivat vagb gur onea, orpnhfr vg unf abg lrg unq gvzr gb erprvir gur vasbezngvba gung gur bgure raq bs gur cbyr unf uvg n jnyy. Gung vasbezngvba pnaabg cebcntngr snfgre guna gur fcrrq bs yvtug. Guvf zrnaf gurer jvyy or n crevbq qhevat juvpu gur onpx raq bs gur cbyr pbagvahrf gb zbir nybat jvgu Nyvpr juvyr gur sebag raq unf orra fgbccrq ol gur jnyy. Gur cbyr qrsbezf, nyybjvat vg gb svg vagb gur onea ol gur gvzr Obo fuhgf gur qbbe.
Bar cbffvovyvgl vf gb unir n plyvaqevpny fcnprgvzr, jurer fcnpr vf xvaq bs yvxr Cnpzna. N crefba geniryyvat ng havsbez irybpvgl va nal qverpgvba jvyy neevir onpx gb jurer fur fgnegrq sebz. Guvf vf n syng fcnprgvzr. Va guvf pnfr, bar gjva pna sbyybj na varegvny cngu gung jvaqf nebhaq gur plyvaqre, juvyr gur bgure gjva sbyybjf n cngu gung tbrf fgenvtug hc gur plyvaqre. Gur nflzzrgel vf qhr gb gbcbybtvpny qvssreraprf orgjrra gur cnguf, abg nppryrengvba.
Gur jbeqyvar bs guvf bowrpg jvyy or n ulcreobyn jvgu gur cubgba’f cngu nf nflzcgbgr.
Gur fgevat jvyy oernx. Sebz gur crefcrpgvir bs gur fcnprfuvcf, gur qvfgnapr orgjrra gurz vapernfrf nf gurl nppryrengr, pnhfvat gur fgevat gb oernx. Vs gur qvfgnapr orgjrra gur fuvcf va gur erfg senzr fgnlf gur fnzr rira gubhtu gur fcnprfuvcf ner nppryrengvat, guvf zrnaf gur qvfgnapr orgjrra gur fuvcf va gurve bja senzr zhfg or vapernfvat (gb pbhagrenpg Yberagm pbagenpgvba). Guvf vf Oryy’f fcnprfuvc cnenqbk, qvfphffrq va terngre qrgnvy urer.
Gubfr gjb cnguf ner vaqrrq flzzrgevp, naq gubfr gjvaf jvyy or gur fnzr ntr jura gurl zrrg. Gur fcnprgvzr vagreiny gurl genirefr jvyy or rknpgyl gur fnzr.
The math of special relativity helps not one whit in solving problems 1, 2 and 4. Problem 3 of course can be answered qualitatively without math, but to indicate with any specificity you will need math.
Moreover, you can be quite proficient with the math of SR and be floored by these. I knew an undergrad who could do acceleration problems yet couldn’t work his way through problem 1, because the course had focused on gamma and neglected spacetime diagrams.
Spacetime diagrams are the math of special relativity. Doing algebra with gammas without insight on how it relates to the Minkowski spacetime is like the proverbial blind men grasping various parts of the elephant. (Why such godawful approaches are still foisted upon students is a mystery to me.)
Some verbal problems on special relativity:
Alice is running very fast holding a long pole. The pole is held parallel to the direction in which she is running. She’s running into a barn with an open door. When the pole is stationary relative to the barn, it does not fit inside it completely. A tiny bit sticks out, preventing the door from being shut. However, since the pole is now in motion, Bob, who is standing by the barn door, sees its length contracted, allowing it to fit completely inside the barn. This means that Bob can wait for Alice to enter the barn and then shut the door. Let’s say he shuts the door as soon as the leading edge of the pole makes contact with the barn wall opposite the door. But from Alice’s perspective, it is the barn’s length that has contracted, not the pole’s. From her perspective, how do you account for the fact that she is able to run the pole completely into the barn so that Bob can shut the barn door?
The usual response to the twin paradox is that the twins’ situation is not symmetrical because the one who leaves Earth must undergo non-inertial motion in order to turn around and return. However, it is possible to reproduce the twin paradox without either twin accelerating, and without moving to curved spacetimes. Can you think of how one might construct this situation?
We all know that according to special relativity it is impossible for a massive object to accelerate until it catches up with a photon. However, it is possible for a massive object to accelerate so that, in its rest frame, its distance from a photon always remains the same. What would the word-line of such an object look like?
We have two spaceships A and B, initially flying at the same uniform velocity. A is flying behind B, and there is a taut string connecting the nose of A to the tail of B. The spaceships both accelerate so that an observer who starts out at rest relative to them will measure the distance between them as unchanging throughout the acceleration. When the ships stop accelerating they are moving relative to the observer. This means the observer should see the string as Lorentz contracted, but, by hypothesis, the distance between the spaceships unchanged. Does this mean the string breaks? Describe what happens from the perspective of the space-ships.
EDIT: Solutions here.
Solutions, or at least sketches of solutions:
Sbe Nyvpr, gur cbyr’f raq uvggvat gur onpx jnyy naq Obo pybfvat gur qbbe jvyy abg or fvzhygnarbhf. Jura gur cbyr uvgf gur onpx raq bs gur jnyy, fbzr bs vg jvyy fgvyy or fgvpxvat bhg bs gur onea. Ohg gung ovg jvyy pbagvahr zbivat vagb gur onea, orpnhfr vg unf abg lrg unq gvzr gb erprvir gur vasbezngvba gung gur bgure raq bs gur cbyr unf uvg n jnyy. Gung vasbezngvba pnaabg cebcntngr snfgre guna gur fcrrq bs yvtug. Guvf zrnaf gurer jvyy or n crevbq qhevat juvpu gur onpx raq bs gur cbyr pbagvahrf gb zbir nybat jvgu Nyvpr juvyr gur sebag raq unf orra fgbccrq ol gur jnyy. Gur cbyr qrsbezf, nyybjvat vg gb svg vagb gur onea ol gur gvzr Obo fuhgf gur qbbe.
Bar cbffvovyvgl vf gb unir n plyvaqevpny fcnprgvzr, jurer fcnpr vf xvaq bs yvxr Cnpzna. N crefba geniryyvat ng havsbez irybpvgl va nal qverpgvba jvyy neevir onpx gb jurer fur fgnegrq sebz. Guvf vf n syng fcnprgvzr. Va guvf pnfr, bar gjva pna sbyybj na varegvny cngu gung jvaqf nebhaq gur plyvaqre, juvyr gur bgure gjva sbyybjf n cngu gung tbrf fgenvtug hc gur plyvaqre. Gur nflzzrgel vf qhr gb gbcbybtvpny qvssreraprf orgjrra gur cnguf, abg nppryrengvba.
Gur jbeqyvar bs guvf bowrpg jvyy or n ulcreobyn jvgu gur cubgba’f cngu nf nflzcgbgr.
Gur fgevat jvyy oernx. Sebz gur crefcrpgvir bs gur fcnprfuvcf, gur qvfgnapr orgjrra gurz vapernfrf nf gurl nppryrengr, pnhfvat gur fgevat gb oernx. Vs gur qvfgnapr orgjrra gur fuvcf va gur erfg senzr fgnlf gur fnzr rira gubhtu gur fcnprfuvcf ner nppryrengvat, guvf zrnaf gur qvfgnapr orgjrra gur fuvcf va gurve bja senzr zhfg or vapernfvat (gb pbhagrenpg Yberagm pbagenpgvba). Guvf vf Oryy’f fcnprfuvc cnenqbk, qvfphffrq va terngre qrgnvy urer.
Jung vs gur gjvaf ner zbivat nebhaq gur plyvaqre ng gur fnzr fcrrq va bccbfvgr qverpgvbaf? Gung’f flzzrgevp.
Gubfr gjb cnguf ner vaqrrq flzzrgevp, naq gubfr gjvaf jvyy or gur fnzr ntr jura gurl zrrg. Gur fcnprgvzr vagreiny gurl genirefr jvyy or rknpgyl gur fnzr.
This is exactly what I need about when I’m starting to need it. Thank you.
The math of special relativity helps not one whit in solving problems 1, 2 and 4. Problem 3 of course can be answered qualitatively without math, but to indicate with any specificity you will need math.
Moreover, you can be quite proficient with the math of SR and be floored by these. I knew an undergrad who could do acceleration problems yet couldn’t work his way through problem 1, because the course had focused on gamma and neglected spacetime diagrams.
I think someone without a detailed grasp of spacetime diagrams should be able to answer problem 1, as long as they know the following principles:
Gur eryngvivgl bs fvzhygnarvgl
Ab fhcreyhzvany fvtanyvat
Spacetime diagrams are the math of special relativity. Doing algebra with gammas without insight on how it relates to the Minkowski spacetime is like the proverbial blind men grasping various parts of the elephant. (Why such godawful approaches are still foisted upon students is a mystery to me.)