DimaMazin Posted September 23, 2015 Author Share Posted September 23, 2015 (edited) Yes, you did. "All parts of object are instantly accelerated to v" and "They are not simultaneously accelerated because forces don't instantly transfer momentum" No, but you eliminated that consideration when you claimed that "All parts of object are instantly accelerated to v" and when I asked you what was happening with the middle of the rocket, you said "middle of the rocket isn't moving" Really my math is considering acceleration and speed after one push. Thank you for idea about speed of force propagation. And so speed of force propagation is faster than c . And reducing object reduces speed of force propagation. Maybe therefore my math contradicts GR, but does it contradict QM ? Edited September 23, 2015 by DimaMazin Link to comment Share on other sites More sharing options...
Mordred Posted September 23, 2015 Share Posted September 23, 2015 (edited) No force propogates at maximum c, if you want good mathematic detail I suggest reading this arxiv paper. Relativistic elasticity of rigid rods and strings http://arxiv.org/abs/1406.0634 It helps to understand the rigid rod if you treat force as being transferred via particle/object to particle/object interactions. Keeping in mind the rod is made up of particles. Edited September 23, 2015 by Mordred 1 Link to comment Share on other sites More sharing options...
swansont Posted September 23, 2015 Share Posted September 23, 2015 Really my math is considering acceleration and speed after one push. Thank you for idea about speed of force propagation. And so speed of force propagation is faster than c . And reducing object reduces speed of force propagation. Maybe therefore my math contradicts GR, but does it contradict QM ? OK, so this nonsense isn't about speed of force propagation. Another confirmation that nobody can make heads or tails of what you're talking about. If you have some idealized material that's infinitely rigid, all parts would accelerate at the same time. Where does your equation for the speed of force propagation come from? No, I'm not asking you to simply repeat it. I'm asking you to derive your result from a basic setup of the problem. Link to comment Share on other sites More sharing options...
Mordred Posted September 23, 2015 Share Posted September 23, 2015 (edited) If it helps a good analogy example is shockwaves from an Earth quake. The mechanical force from the epicentre is transmitted outward at the speed of sound through a medium. Variations in the waves speed through the medium vary depending on the type of wave. https://en.m.wikipedia.org/wiki/Seismic_wave same basic physics apply to force through a medium ie rigid rod. Edited September 23, 2015 by Mordred Link to comment Share on other sites More sharing options...
DimaMazin Posted September 24, 2015 Author Share Posted September 24, 2015 (edited) No force propogates at maximum c, .Agreed. If you have some idealized material that's infinitely rigid, all parts would accelerate at the same time. If you have no such material then what proves that all parts of accelerated and contracted object are accelerated at the same time? Edited September 24, 2015 by DimaMazin Link to comment Share on other sites More sharing options...
swansont Posted September 24, 2015 Share Posted September 24, 2015 If you have no such material then what proves that all parts of accelerated and contracted object are accelerated at the same time? This is physics. We do idealized thought experiments all the time. Link to comment Share on other sites More sharing options...
DimaMazin Posted September 27, 2015 Author Share Posted September 27, 2015 This is physics. We do idealized thought experiments all the time. Even when they are accelerated at the same time then they are initially accelerated to different speeds. Physics should be idealized only by math. Link to comment Share on other sites More sharing options...
swansont Posted September 27, 2015 Share Posted September 27, 2015 Even when they are accelerated at the same time then they are initially accelerated to different speeds. Physics should be idealized only by math. Then let's see the math. From the initial setup — not just writing down what you assert is the answer. Show your work. Link to comment Share on other sites More sharing options...
DimaMazin Posted September 27, 2015 Author Share Posted September 27, 2015 (edited) Then let's see the math. From the initial setup — not just writing down what you assert is the answer. Show your work. I think necessary distance for forces balance is reducing therefore forces of counteraction have different shifts. Counteraction force of tail has blueshift, counteraction force of head has redshift to motion direction. Edited September 27, 2015 by DimaMazin -1 Link to comment Share on other sites More sharing options...
swansont Posted September 27, 2015 Share Posted September 27, 2015 I think necessary distance for forces balance is reducing therefore forces of counteraction have different shifts. Counteraction force of tail has blueshift, counteraction force of head has redshift to motion direction. That's not math, and forces have red or blue shifts? Link to comment Share on other sites More sharing options...
DimaMazin Posted September 29, 2015 Author Share Posted September 29, 2015 That's not math, and forces have red or blue shifts? Let's consider example when you try to accelerate head first. Head,middle and tail are connected by strong interaction. You,trying to accelerate head, increase distance between the head and the middle. Increase of gap increases strong interaction.Also acceleration increases speed and length contraction, which increase gap too. It creates more strong interaction even between the middle and the head.When the strong interaction force is bigger than acceleration force for the head then the strong interaction force accelerates the middle faster than the head acceleration. The tail accepts sufficient force for acceleration and reducing of distance to the head from bigger force of strong interaction. Link to comment Share on other sites More sharing options...
swansont Posted September 29, 2015 Share Posted September 29, 2015 Let's consider example when you try to accelerate head first. Head,middle and tail are connected by strong interaction. You,trying to accelerate head, increase distance between the head and the middle. Not for an ideal material that is infinitely rigid. For a real material, yes, but that will depend on the properties of the material, which don't appear in your equations, so you can't be taking into account this effect. Further, when I asked about this, you said that the middle part was not moving at a different velocity. Your examples are not consistent with each other. It also is probably not helpful to switch examples in the middle of the discussion. It can only add to how muddled the conversation will be. Let's stick to accelerating the tail. Increase of gap increases strong interaction. "Strong interaction" has a meaning in physics. How does the strong interaction apply here? When the strong interaction force is bigger than acceleration force for the head then the strong interaction force accelerates the middle faster than the head acceleration. Unless the external force changes, this will not happen. The elasticity force will behave according to Hooke's law, F = -kx (or similar behavior) The material will stretch until the forces are equal. Then there is no relative acceleration, so there is no further stretching. You never get to the point where the elasticity will get larger without changing the external force. Link to comment Share on other sites More sharing options...
DimaMazin Posted October 10, 2015 Author Share Posted October 10, 2015 Not for an ideal material that is infinitely rigid. Let's consider math of length contraction of ideal material. Forces of object at v are equal on contracted distance. Therefore acceleration causes forces of length contraction to mathematical middle of the object. For example masses of tail,middle and head are equal. Therefore: Force of head acceleration = F Force of length contraction for head at start of motion= -F Force of tail acceleration=F Force of length contraction of object for tail=F Therefore force of tail motion at start of motion=2F Force of head motion at start of motion=0 Link to comment Share on other sites More sharing options...
swansont Posted October 10, 2015 Share Posted October 10, 2015 Let's consider math of length contraction of ideal material. Forces of object at v are equal on contracted distance. Therefore acceleration causes forces of length contraction to mathematical middle of the object. For example masses of tail,middle and head are equal. Therefore: Force of head acceleration = F Force of length contraction for head at start of motion= -F Force of tail acceleration=F Force of length contraction of object for tail=F Therefore force of tail motion at start of motion=2F Force of head motion at start of motion=0 Nope. Not making any sense. Length contraction is not caused by a force, so what is force of length contraction? Further, the force is being exerted in the object's frame, so there is no length contraction present. Link to comment Share on other sites More sharing options...
DimaMazin Posted October 10, 2015 Author Share Posted October 10, 2015 (edited) Nope. Not making any sense. Length contraction is not caused by a force, so what is force of length contraction? Further, the force is being exerted in the object's frame, so there is no length contraction present. Do you think that object forces aren't in balance on defined distance?And the distance isn't changed relative frameS? https://en.wikipedia.org/wiki/Reaction_%28physics%29 Force of length contraction is caused by reaction of forces connection to past distance which try to break off the object with force of acceleration. Edited October 10, 2015 by DimaMazin Link to comment Share on other sites More sharing options...
swansont Posted October 10, 2015 Share Posted October 10, 2015 Do you think that object forces aren't in balance on defined distance?And the distance isn't changed relative frameS? In balance with respect to what? If an object is accelerating, there had better be a net force on it. "distance isn't changed relative frameS?" Sorry, don't understand what this means. In frame S there is no contraction. Unless you have changed things, that's the frame of the rocket. Other frames will observe contraction, but that's not what we're discussing. I thought we were talking about a rocket in its own frame. Link to comment Share on other sites More sharing options...
DimaMazin Posted October 10, 2015 Author Share Posted October 10, 2015 (edited) In balance with respect to what? If an object is accelerating, there had better be a net force on it. "distance isn't changed relative frameS?" Sorry, don't understand what this means. In frame S there is no contraction. Unless you have changed things, that's the frame of the rocket. Other frames will observe contraction, but that's not what we're discussing. I thought we were talking about a rocket in its own frame. net force=2F for tail +F for middle + 0 for head at start of motion =F for tail+F for midde +F for head at end of acceleration Edited October 10, 2015 by DimaMazin Link to comment Share on other sites More sharing options...
swansont Posted October 10, 2015 Share Posted October 10, 2015 net force=2F for tail +F for middle + 0 for head at start of motion =F for tail+F for midde +F for head at end of acceleration Why? Link to comment Share on other sites More sharing options...
DimaMazin Posted October 10, 2015 Author Share Posted October 10, 2015 Why? Net force can't simultaneously accelerate and contract all parts of object even mathematically. Object at v has contracted distance of own forces balance. Therefore preservation of distance causes force of reaction against the distance.We can call the force by force of length contraction. Link to comment Share on other sites More sharing options...
swansont Posted October 10, 2015 Share Posted October 10, 2015 Net force can't simultaneously accelerate and contract all parts of object even mathematically. It would be nice if you'd present the math that shows this, but I'm tired of going around in circles asking for it. Link to comment Share on other sites More sharing options...
DimaMazin Posted October 11, 2015 Author Share Posted October 11, 2015 (edited) It would be nice if you'd present the math that shows this, but I'm tired of going around in circles asking for it. If you don't understand relativity then you should consider usual analogue. For example you push or pull very long and hot metal rod. Cold wind cools the rod. Then force of length contraction of the rod increases force of acceleration of tail of the rod and reduces force of acceleration of head of the rod. Do you deny even this? Edited October 11, 2015 by DimaMazin Link to comment Share on other sites More sharing options...
swansont Posted October 11, 2015 Share Posted October 11, 2015 If you don't understand relativity then you should consider usual analogue. For example you push or pull very long and hot metal rod. Cold wind cools the rod. Then force of length contraction of the rod increases force of acceleration of tail of the rod and reduces force of acceleration of head of the rod. Do you deny even this? Length contraction is not a force; cooling the rod is not part of relativity nor is it a good analogy. The rod is in its own frame, and as such it does not contract. My understanding of relativity is not the issue here. It's your inability to articulate your argument in a rigorous fashion. Link to comment Share on other sites More sharing options...
DimaMazin Posted October 11, 2015 Author Share Posted October 11, 2015 Length contraction is not a force; cooling the rod is not part of relativity nor is it a good analogy. The rod is in its own frame, and as such it does not contract. My understanding of relativity is not the issue here. It's your inability to articulate your argument in a rigorous fashion. Then all is in own frame and nothing is contracted. I don't understand such science . Link to comment Share on other sites More sharing options...
Strange Posted October 11, 2015 Share Posted October 11, 2015 I don't understand such science . That is the only thing you have said that I understood. It also seems to be factually correct. 2 Link to comment Share on other sites More sharing options...
DimaMazin Posted October 12, 2015 Author Share Posted October 12, 2015 That is the only thing you have said that I understood. It also seems to be factually correct. I thought observer makes math of moving object relative to own frame.And when observer can't make the math then he/she isn't a scientist. Link to comment Share on other sites More sharing options...
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